Buffer questions
A solution containing 50.00 mL of 0.1800 M NH3 (Kb = 1.77 x 10-5) is being titrated with 0.3600 M HCl. Calculate the pH: e. After the addition of 26.00 mL of HCl
1) Let us determine the new molarity of the HCl: M1V1 = M2V2 (0.36) (1.00 mL) = (x) (76.00 mL) x = 0.0047368 M (I kept some guard digits.) 2) Calculate the pH in the usual manner for a strong acid: pH = -log [H+] pH = 2.324
Determine how you would prepare 1.00 L of this buffer starting with 0.100 M HA, 0.100 M NaOH and water where the total concentration of HA plus NaA is 0.010 M?
The total moles of HA and NaA = 0.010 mol. This comes from the total molarity (0.0100 M) times the final volume of the solution. 5.000 = 4.700 + log [x / (0.01 - x)] x = 0.00667 mole of the base (the A¯) a) x represents the moles of the NaA (the salt) in the 1.00 L of solution b) 0.010 - x = 0.00333; this is the moles of HA in 1.0 L of solution. 1) take 100. mL of 0.100 M HA. This represents 0.0100 mole of HA. 2) Add 66.7 mL of 0.100 M NaOH solution. This is 0.0067 mole of NaOH. 3) dilute to 1.00 L with water.
Determine the pH of a solution prepared by dissolving 0.35 mole of ammonium chloride in 1.0 L of 0.25 M aqueous ammonia. Kb for ammonia equals 1.77 x 10-5
This is a buffer solution, weak base (NH3) & weak base salt (NH4+) in solution at the same time. pH = pKa + log [0.25 / 0.35] 1.00 x 10^-14 = (Ka) (1.77 x 10^-5)=> 5.65 x 10^-10 pH = 9.248 + log [0.25 / 0.35]= 9.10
If an acetate buffer solution was going to be prepared by neutralizing HC2H3O2 with 0.10 M NaOH, what volume (in mL) of 0.10 M NaOH would need to be added to 10.0 mL of 0.10 M HC2H3O2 to prepare a solution with pH = 5.50?
initial moles of acetic acid: (0.010 L) (0.1 mol/L) = 0.001 mol 5.50 = 4.752 + log (x / (0.001 - x)) x = 8.5 x 10-4 moles volume of NaOH required: 8.5 x 10-4 mol divided by 0.1 M = 8.5 x 10-3 L
Calculate the pH when 50.0 mL of 0.180 M NH3 is mixed with 5.00 mL of 0.360 M HBr. (The Kb of ammonia is 1.77 x 10-5.)
1) (0.180 mol/L) (0.0500 L) = 0.00900 mol ammonia (0.360 mol/L) (0.0050 L) = 0.00180 mol of HBr 2) ammonia : 0.00900 - 0.00180 = 0.00720 mol NH3 that reacts with HBr produces ammonium ion In the solution will be 0.00180 mole of NH4+ pH = 9.248 + log (0.00720 / 0.00180)= 9.850
How many mL of 0.75 M HCl must be added to 120 mL of 0.90 M sodium formate to make a buffer of pH = 4.00? pKa of formic acid = 3.75
(0.90 mol/L) (0.12 L) = 0.108 mol <--- formate is base The HCl will protonate formate, make formic acid. 4.00 = 3.75 + log (0.108 - x) / x x = 0.038873 mol Volume HCl: 0.038873 mol / 0.75 mol/L = 0.05183 L pH = 3.75 + log (0.069127 / 0.038873)=4
A solution containing 50.00 mL of 0.1800 M NH3 (Kb = 1.77 x 10-5) is being titrated with 0.3600 M HCl. Calculate the pH: a. initially
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How much sodium formate (HCOONa, 68.0069 g/mol) do you need to add to 400. mL of 1.00 M formic acid for a pH 3.500 buffer. Ka = 1.77 x 10¯4
3.500 = 3.752 + log (x/1) <--- 1.00 M formic acid x = 0.560 M, molarity of required formate MV = g/molar mass (0.560 mol/L) (0.400 L) = x / 68.0069 g/mol x = 15.2 g
You need to prepare a buffer solution of pH 3.972 from 10.0 mL of 0.335 M solution of a weak acid whose pKa is 3.843. What volume of 0.385 M NaOH would you need to add?
3.972 = 3.843 + log [A¯][HA] [A¯] = 1.346[HA] We also know that [A¯] + [HA] = 0.335 M. 1.346[HA] + [HA] = 0.335 [HA] = 0.143 M & [A¯] = 0.192 M Calculate moles of A¯=NaOH moles (0.192 mol/L) (0.010 L) = 1.92 x 10¯3 mol A¯= OH¯ volume of NaOH required: 1.92 x 10¯3 mol / 0.385 M = 4.99 x 10¯3 L
What volumes of 0.100 M acetic acid and 0.100 M sodium acetate would be required to produce 1.00 L of buffer at pH 4.000? (pKa = 4.752)
4.000 = 4.752 + log (Ac¯/HAc) [HAc]/[Ac¯] = 5.6493 5.64937x + x = 1000 x = 150.39 mL of acetate plus 849.61 mL acetic acid
Calculate the ratio of the concentration of acetic acid and acetate required in a buffer system at a pH of 4.208 (the pKa of acetic acid equals 4.752).
4.208 = 4.752 + log (base / acid) log (base / acid) = -0.544 base / acid = 0.285759 acid / base ratio is 3.5 to 1
What mass of HCl would need to be added to a 250. mL solution containing 0.500 M NaC2H3O2 and 0.500 M HC2H3O2, to make the pH = 4.25? Ka of HC2H3O2 is 1.77 x 10-5.
4.25 = 4.752 + log [base / acid] [base/acid] = 0.314775 0.125 - x / 0.125 + x = 0.314775 x = 0.0651466 mol (0.0651466 mol) (36.5 g/mol) = 2.38 g
A solution containing 50.00 mL of 0.1800 M NH3 (Kb = 1.77 x 10-5) is being titrated with 0.3600 M HCl. Calculate the pH: b. After the addition of 5.00 mL of HCl
Determine moles of NH3 and HCl before mixing: NH3: (0.1800 mol/L) (0.0500 L) = 0.00900 mol HCl: (0.3600 mol/L) (0.0050 L) = 0.00180 mol Determine moles of NH3 and NH4+ after mixing: NH3: 0.00900- 0.00180= 0.00720 mol NH3 rem NH4+: 0.00180 mol produced pH = 9.248 + log [(0.0072/0.055) / (0.0018/0.055)] pH = 9.850
Determine the pH of the buffer made by mixing 0.0300 mol HCl with 0.0500 mol CH3COONa in 2.00 L of solution. The Ka of acetic acid is 1.77 x 10¯5.
H+ ion from the HCl will protonate the acetate ion rxn: H+ + CH3COO¯ ---> CH3COOH CH3COOH ---> 0.0300 mol is made CH3COO¯-> 0.0500 - 0.0300 = 0.0200 mol remains pH = 4.752 + log [0.0200 / 0.0300]=4.928
1.00 L of a solution containing 0.0500 mole of HOAc and 0.100 mole of NaOAc is prepared. Ignore the autoionization of water for the purposes of this problem. The Ka of HAc equals 1.77 x 10-5. (a) Calculate the numerical value of the reaction quotient, Q for the initial condition.
HAc ⇌ H+ + Ac¯ Write the reaction quotientand solve for Q: Qa = ([H+]o [Ac¯]o) / [HAc]o Qa = [(1.00 x 10¯7) (0.100)] / 0.0500 Qa = 2.00 x 10¯7 value for [H+] derived from autoionization water.
pKa for phenophthalein is 9.3 at room temp. a) Calculate ratio of its anionic form to acid form at pH 8.2 and at pH 10. b) Using these values, explain the colour change within this pH range.
(a): At pH = 8.2: 8.2 = 9.3 + log (base form / acid form) log (base form / acid form) = -1.1 ratio of base form to acid form = 0.0794 to 1 (call it 8 to 100) 2) At pH = 10.: 10 = 9.3 + log (base form / acid form) ratio of base to acid = 5.01 to 1 (call it 500 to 100 --- b): it's the anionic (or base) form that is colored pink. The acid form is colorless. At pH = 8.3: the pink form is in the minority at this pH. For every 100 acid (colorless) forms, there are only 8 base (pink) forms. 3) At pH = 10.: the colorless form is in the minority. For every 100 acid (colorless) forms present, there are now 500 base (pink) forms present. 4) This means: From pH = 8.3 to pH = 10., there has been a 6250% increase in pink forms (from 8:100 to 500:100).
20.0 cm3 of 1.00 mol HNO2 and 40.0 cm3 of 0.500 NaNO2 are mixed. What is the pH of the resulting solution? pKa of nitrous acid is 3.34
1) Common ion effect....The presence of a common ion, in this case the nitrite ion, will reduce the ionization of the weak acid HNO2 2) HNO2 is diluted by the addition of NaNO2 sln 20 mL HNO2 x 1.0vM / 60 mL = 0.33vM HNO2 3) The [NO2¯ ] is 2/3 of its original concentration. 2/3 x 0.5 = 0.33M NO2¯ 4) Since the [HNO2 ]=[NO2¯] ,cancel out, [H+]= Ka, pH=pKa. Ka = [H+][NO2¯] / [HNO2] 4.57 x 10¯4 = [(x)(0.33M)] / 0.33M] x = 4.57 x 10¯4 = [H+] 5) pH = -log(4.57x10¯4)=3.34
A solution containing 50.00 mL of 0.1800 M NH3 (Kb = 1.77 x 10-5) is being titrated with 0.3600 M HCl. Calculate the pH: c. After the addition of a total volume of 12.50 mL HCl
1) Determine moles of NH3 and HCl before mixing: NH3: (0.1800 mol/L) (0.0500 L) = 0.00900 mol HCl: (0.3600 mol/L) (0.01250 L) = 0.00450 mol Determine moles of NH3 and NH4+ after mixing: NH3: 0.00900- 0.00450= 0.00450 mol/rem NH4+: 0.00450 mol produced pH = 9.248 + log (0.0045 / 0.0045)= 9.248
A solution containing 50.00 mL of 0.1800 M NH3 (Kb = 1.77 x 10-5) is being titrated with 0.3600 M HCl. Calculate the pH: d. After the addition of a total volume of 25.00 mL of HCl
1) Determine moles of NH3 and HCl before mixing: NH3: (0.1800 mol/L) (0.0500 L) = 0.00900 mol HCl: (0.3600 mol/L) (0.02500 L) = 0.00900 mol 2) Determine moles of ammonia and ammonium ion after mixing: NH3: 0.00900 - 0.00900 m= zero mol NH3 rem NH4+: 0.00900 mol produced The actual molarity is (0.00900 mol / 0.0750 L)=0.120 . pH =5.084
You need to prepare an acetate buffer of pH 6.420 from a 0.664 M acetic acid solution and a 2.50 M KOH solution. If you have 975 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.42? The pKa of acetic acid is 4.752.
1) Moles of acetic acid: (0.664 mol/L) (0.975 L) = 0.6474 mol 2) pH = pKa + log [base / acid] 6.420 = 4.752 + log [x / (0.6474 - x)] 3) x = 0.63379 mol Acetate and KOH are in a 1:1 stoichiometric ratio, so this is the required number of moles of KOH 4) Volume of KOH needed: 0.63379 mol / 2.50 mol/L = 0.253516 L 254 mL seems like a reasonable answer try our calculated values and see what happens: pH = 4.752 + log (0.63379 / 0.01361) = 6.420
You have 0.500 liter of an acetic acid buffer (0.800 M total) at maximum buffering capacity. To it, you add 0.100 mole of HCl. What is the new pH
1) Moles present in solution: moles HAc ---> 0.400 mol moles acetate ---> 0.400 mol Max buf cap, when HA and its A- in 1:1 molar ratio 2) moles HAc ---> 0.400 + 0.100 = 0.500 moles acetate ---> 0.400 - 0.100 = 0.300 3) pH = 4.752 + log (0.300 / 0.500)= 4.530
Calculate the pH when 25.0 mL of 0.200 M acetic acid is mixed with 35.0 mL of 0.100 M NaOH.
1)(0.200 mol/L) (0.0250 L) = 0.00500 mol of aceticA (0.100 mol/L) (0.0350 L) = 0.00350 mol of NaOH 2) Determine the moles remaining after reaction acetic acid (in excess) ---> 0.00500 mol - 0.00350 mol = 0.00150 mol The acetic acid that reacts with NaOH produces sodium acetate.In the solution will be 0.00350 mol of acetate anion pH = 4.752 + log (0.00350 / 0.00150)= 5.120
Calculate the pH of the solution that results from the addition of 0.040 moles of HNO3 to a buffer made by combining 0.500 L of 0.380 M HC3H5O2 (Ka = 1.30 x 10¯5) and 0.500 L of 0.380 M NaC3H5O2. Assume addition of the nitric acid has no effect on volume.
1a) The HNO3 reduce NaC3H5O2: (0.380 mol/L) (0.500 L) = 0.190 mol of NaC3H5O2 0.190 - 0.040 = 0.150 mol NaC3H5O2 remaining 1b) The reaction in 1a increase HC3H5O2: increase by the same amount of the decrease 0.190 mol + 0.040 mol = 0.230 mol of HC3H5O2 2) pH = 4.886 + log (0.150 / 0.230)=4.7
You desire to create a solution with a pH of 3.26. If you add 0.577 moles of HF to 1.00 L of solution, how many moles of NaF should you add? Ka of HF: 7.2 x 10¯4
3.26 = 3.14 + log (x / 0.577) x = 0.76 mol
If you begin with 48.2 mL of a 0.171 M solution of HNO2, how many grams of NaNO2 would you have to add to the solution for a pH of 3.32?
3.32 = 3.40 + log (x / 0.171) x = 0.14223 M (NaNO2 g=? required 48.2 mL of a 0.14223 M sln. ) MV = mass / molar mass (0.14223 mol/L) (0.0482 L) = mass / 68.995 g/mol mass = 0.473 g
You need to prepare a buffer solution of pH 4.178 from 25.0 mL of 0.282 M solution of a sodium salt of a weak acid, NaA where the pKa of the weak acid HA is 4.270. What volume of 0.329 M HCl would you need to add?
4.178 = 4.270 + log [base][acid] The A-of weak acid is base &is the acid. Adding HCl will turn some of the A¯ into HA. A-: (0.282 mol/L) (0.025 L) = 0.00705 mol 4.178 = 4.270 + log 0.00705 - xx 0.00705 - x ---> A¯ go down when reacting W/ HCl x ---> that's the amount of HA produced log 0.00705 - xx = -0.092 x = 0.00389697 mol < moles of HA in base/acid 1:1 molar ratio; also the moles of HCl we need 0.00389697 mol / 0.329 mol/L = 0.0118449 L
What mass of HCl would need to be added to a 250. mL solution containing 0.500 M NaC2H3O2 and 0.500 M HC2H3O2, to make the pH = 4.25? Ka of HC2H3O2 is 1.77 x 10¯5
4.25 = 4.752 + log [base / acid] [base / acid] = 0.314775, get it to 4.25 (0.125 - x) / (0.125 + x)= 0.31477; from (0.5 mol/L times 0.25 L) x = 0.0646394 mol 0.0646394 mol times 36.4609 g/mol = 2.36 g
A buffer with a pH of 4.30 contains 0.33 M of sodium benzoate and 0.26 M of benzoic acid. What is the concentration of H+ in the solution after the addition of 0.058 mol of HCl to a final volume of 1.6 L?
4.30 = pKa + log(0.33/0.26) Determine moles after addition of HCl: 0.330 - 0.058 = 0.272 mole sodium benzoate 0.260 + 0.058 = 0.318 mole benzoic acid pH = 4.196 + log(0.272/0.318)=4.128 [H+] = 10-4.128 = 7.745 x 10-5 M
A buffer with a pH of 4.31 contains 0.31 M of sodium benzoate and 0.24 M of benzoic acid. What is the concentration of [H+] in the solution after the addition of 0.050 mol of HCl to a final volume of 1.3 L? Assume that any contribution of HCl to the volume is negligible.
4.31 = pKa + log (0.31 / 0.24)=>4.20 3)how much benzoate is reacted w/ HCl. (0.31 mol/L) (1.3 L) = 0.403 mol 0.403 mol - 0.050 mol = 0.353 mol 4)Benzoic acid went up by the same 0.050 mol: (0.24 mol) (1.3 L) = 0.312 mol 0.312 mol + 0.050 mol = 0.362 mol 5)pH = 4.20 + log (0.353 / 0.362)= 4.189 [H+] = 10¯pH = 10¯4.189 = 0.000065 M
We desire to make a pH 5.000 buffer and we choose a weak acid (let's call it HA) with a pKa of 4.700. Starting with 0.100 M each HA and NaA, we desire to make 100. mL buffer solution.
5.000 = 4.700 + log [A¯] / [HA] [A¯] / [HA] = 2.00 moles of A¯ = (0.100 mol / L) (LA¯) moles of HA = (0.100 mol / L) (LHA) --------- [(0.100 mol/L) (LA¯)] / [(0.100 mol/L) (LHA)] = 2.00 (LA¯) / (LHA) = 2.00 -------- let LA¯ = x therefore LHA = 0.1 - x x / (0.1 - x) = 2.00 x = 0.667 L We require 66.7 mL of NaA and 33.3 mL of HA to make our pH 5 buffer.
A beaker with 100.0 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1000 M. A student adds 7.300 ml of a 0.3600 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.752.
5.000 = 4.752 + log (x / 1) x = 1.77 Base:salt ratio is 1.77:1 0.0100 mole of total solute: let x = moles of acid 1.77x + x = 0.01 total moles of solute x = 0.00361 moles (this is the acid) base = 0.00639 mol (0.360 mol/L) (0.00730 L) = 0.002628 mol HA STR base: 0.00639 - 0.002628 = 0.003762 mol acid: 0.00361 + 0.002628 = 0.006238 mol pH = 4.752 + log (0.003762/0.006238)= 4.532
A beaker with 175 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.40 mL of a 0.300 M HCl solution to the beaker. What is the new pH? The pKa of acetic acid is 4.752.
5.000 = 4.752 + log [x / (0.1 - x)] x = 0.0639 M <--- [A-] in pH 5 buffer 0.1 - 0.0639 = 0.0361 [HA] acetate :(0.0639 mol/L) (0.175 L) = 0.0111825 mol acetic acid :(0.0361 mol/L) (0.175 L) = 0.0063175 mol HCl :(0.300 mol/L) (0.00840 L) = 0.00252 mol The HCl react w/ acetate => acetic acid. acetate :0.0111825 - 0.00252 = 0.0086625 mol acetic aid:0.0063175 + 0.00252 = 0.0088375 mol pH = 4.752 + log (0.0086625 / 0.0088375)= 4.743
You need to produce a buffer solution that has a pH of 5.270. You already have a solution that contains 10.0 mmol (millimoles) of acetic acid. How many millimoles of sodium acetate will you need to add to this solution? The pKa of acetic acid is 4.752.
5.270 = 4.752 + log (x / 10) log (x / 10) = 0.518 x / 10 = 3.296 x = 33.0 millimoles of sodium acetate
You need to prepare an acetate buffer of pH 5.83 from a 0.642 M acetic acid solution and a 2.31 M KOH solution. If you have 975 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.830? The pKa of acetic acid is 4.752.
5.830 = 4.752 + log [base / acid] The acetic acid amount is this: 0.62595 - x The 0.62595 mol :(0.642 mol/L) (0.975 L) = 0.62595 5.830 = 4.752 + log [x / (0.62595 - x)] x = 0.57768 mol of acetate required Because of the 1:1 molar ratio between OH¯ consumed and acetate produced 0.57768 mol divided by 2.31 mol/L = 0.250078 L
Calculate the volume (in mL) of 0.170 M NaOH that must be added to 311 mL of 0.0485 M HA (a generic weak acid) to give the solution a pH of 7.55. The pKa of HA = 7.18.
7.55 = 7.18 + log (A¯) / (HA minus A¯) moles HA ---> (0.0485 mol/L) (0.311 L) = 0.0150835 7.55 = 7.18 + log [(x) / (0.0150835 - x)] x = 0.0105732 moles of A¯ required 0.0105732 mol / 0.170 mol/L = 0.0621953 L NaOH
How many grams of NH4Cl need to be added to 1.50 L of 0.400 M ammonia in order to make a buffer solution with pH of 8.58? Kb for ammonia is 1.77 x 10¯5
8.58 = pKa + log [base / acid] 1.00 x 10¯14 = (Ka) (1.77 x 10¯5) Ka = 5.6497 x 10¯10 pKa = - log 5.6497 x 10¯10 = 9.24797 8.58 = 9.24797 + log [base / acid] (0.400 mol/L) (1.50 L) = 0.600 mol NH3 Add it in: 8.58 = 9.24797 + log [0.6 / x] x = 2.7933234 mol 2.7933234 mol times 53.4916 g/mol = 149.42 g molarity of the NH4Cl: 2.7933234 mol / 1.50 L
How many grams of dry NH4Cl need to be added to 2.40 L of a 0.800 M solution of ammonia to prepare a buffer solution that has a pH of 8.90? Kb for ammonia is 1.77 x 10¯5.
8.90 = 9.248 + log [0.800 / x] x = 1.782748 M ( required molarity of the NH4Cl) (1.782748 mol/L) (2.40 L) = mass / 53.4916 g/mol mass = 228.8689 g
Aspirin has a pKa of 3.4. What is the ratio of A¯ to HA in: (a) the blood (pH = 7.4) (b) the stomach (pH = 1.4)
A) 7.4 = 3.4 + log ([A¯] / [HA]) 10^4 = [A¯] / [HA] B) 1.4 - 3.4 = - 2.0 = log ([A¯] / [HA])= 0.01
(b) Calculate the pH after adding 50.0 mL of a 1.00 M NaOH solution.
A)First calculate the amounts before the addition of the NaOH: HCOOH ---> (0.700 mol/L) (0.500 L) = 0.350 mol HCOONa ---> (0.500 mol/L) (0.500 L) = 0.250 mol B)Determine the moles of NaOH: NaOH ---> (1.00 mol/L) (0.0500 L) = 0.0500 mol C) NaOH reacts in a 1:1 molar ratio with HCOOH: HCOOH ---> 0.350 mol minus 0.0500 mol = 0.300 HCOONa ---> 0.250 mol + 0.0500 mol = 0.300 mol Calculate the new pH: pH = 3.77 + log [0.300 / 0.300] pH = 3.77 + log 1 pH = 3.77
Calculate the pH after 0.020 mol of HCl is added to 1.00 L of 0.10 M NaC3H5O2. Ka = 1.3 x 10-5.
Before the HCl reacts: NaC3H5O2 ---> (0.10 mol/L) (1.0 L) = 0.10 mol HCl ---> 0.02 mol <--- it's the limiting reagent After the HCl reacts NaC3H5O2 ---> 0.10 mol minus 0.02 mol = 0.08 mol HC3H5O2 ---> 0.02 mol pKa=-log 1.3 x 10-5 = 4.886 pH = 4.886 + log (0.08 / 0.02)= 5.49
0.1 mole of CH3NH2 (Kb = 5 x 10¯4) is mixed with 0.08 mole of HCl and diluted to one liter. What will be the H+ concentration?
CH3NH2 + H+ ---> CH3NH3+ 2) When 0.1 mole of CH3NH2 and 0.08 mole of HCl react, this is what remains after the reaction: 0.02 mol CH3NH2 -- 0.08 mol CH3NH3+ 3) pKa = 14 - 3.30103 = 10.69897 4) pH = 10.69897 + log (0.02 / 0.08) = 10.09691 [since it's in 1 liter, these are the molarities] [H+] = 10¯pH = 10¯10.09691 = 8 x 10¯11 M
A solution containing 0.0158 M a diprotic acid with the formula H2A and 0.0226 M of its salt Na2A. The K2 values for the acid are 1.20 x 10¯2 (Ka2) and 5.37 × 10¯7 (Ka2). What is the pH of the solution?
H2A ⇌ HA¯ + H+ Ka1 HA¯ ⇌ A2¯ + H+ Ka2 2) Reverse the second equilibrium: H2A ⇌ HA¯ + H+ Ka1 A2¯ + H+ ⇌ HA¯ 1/Ka2 3) Add the two chemical equations: H2A + A2¯ ⇌ 2HA¯ K' = Ka1 / Ka2 4) Solve for K': K' = 1.20 x 10¯2 / 5.37 × 10¯7 = 22346 5) What we now do is treat this as a limiting reagent problem:H2A + A2¯ ⇌ 2HA¯ Assume 1.00 L of solution is present. moles H2A ---> 0.0158 moles A2¯ ---> 0.0226 The 1:1 stoichiometry of H2A reacting with A2¯ means H2A (the lower amount at 0.0158) is the limiting reagent. Determine amount of HA¯ and H2A in solution: A2¯ ---> 0.0226 - 0.0158 = 0.0068 HA¯ ---> (2) (0.0158) = 0.0316 Reaction stoichiometry: two HA¯ produced for every one H2A used. 7) Calculate the pH as a buffer centered around Ka2 pH = 6.270 + log (0.0068/0.0316) = 5.603
What is the pH of the resulting solution if 40. mL of 0.432 M methylamine, CH3NH2, is added to 15 mL of 0.234 M HCl? Assume that the volumes of the solutions are additive. Ka = 2.70 × 10-11 for CH3NH3+.
HCl ---> (0.234 mol/L) (0.015 L) = 0.00351 mol CH3NH2 ---> (0.432 mol/L) (0.040 L) = 0.01728 mol 100% of the HCl is used up. methylamine ---> 0.01728-0.00351= 0.01377 mol methylammonium ion ---> 0.00351 mol pKa CH3NH3+ = -log 2.70 x 10-11 = 10.5686 pH = 10.5686 + log (0.01377 / 0.00351) = 11.1622
In 1.00 L of solution, 0.529 mole of HNO2 is added to 0.246 mole of NaOH. (Nitrous acid has a Ka of 4.0 x 10¯4.) What is the final pH?
HNO2 to react w/ NaOH to form some NaNO2: moles HNO2 remaining > 0.529 - 0.246 = 0.283 mol moles NaNO2 formed > 0.246 mol HNO2 and NaOH react 1:1 molar ratio form NaNO2 pH = 3.40 + log (0.246 / 0.283]=3.34
1.00 L of a solution containing 0.0500 mole of HOAc and 0.100 mole of NaOAc is prepared. Ignore the autoionization of water for the purposes of this problem. The Ka of HAc equals 1.77 x 10-5. (c) Calculate to 3 significant digits the pH of this solution.
Ka = ([H+] [Ac¯]) / [HAc] 1.77 x 10¯5 = [(x) (0.100)] / 0.0500 x = 8.85 x 10¯6 M pH = -log 8.85 x 10¯6 = 5.053
Calculate the ratio of CH3NH2 to CH3NH3Cl required to create a buffer with pH = 10.14
Ka = 1.00 x 10-14 / 4.4 x 10-4 = 2.27 x 10^-11 Methyl CH3NH3+ + H2O ⇌ H3O+ + CH3NH2 10.14 = 10.644 + log (base/acid) log (base/acid) = -0.504 base/acid ratio = 0.313
A buffer is prepared containing 1.00 M acetic acid and 1.00 M sodium acetate. What is its pH?
Ka of acetic acid is 1.77 x 10¯5 pKa = -log Ka = -log 1.77 x 10¯5 = 4.752 pH = 4.752 + log( 1.00/1.00)=4.752
What would the concentration of sodium formate (NaCOOH) be in 0.00750 M formate buffer at pH 4.358?
Ka of formic acid= 1.77 x 10¯4 formic acid: 0.00750 - x sodium formate: x 4.358 = 3.752 + log [x / (0.00750 - x)] x = 0.00601 M pH = 3.752 + log [0.00601 / 0.00149]= 4.358
What is the [A-]/[HA] ratio when a weak acid is in a solution with the pH one unit below its pKa
Let use set the pKa to be K and the pH to be K - 1 pH = pKa + log ([A¯]/[HA]) K - 1 = K + log ([A¯]/[HA]) -1 = log ([A¯]/[HA]) 1/10 = ([A¯]/[HA])
200.0 mL of an acetate/acetic acid buffer is 0.100 M in total molarity and has a pH of 5.000. After 6.30 mL of 0.490 M HCl is added, what is the new pH?
MV = (0.100 mol/L) (0.2000 L) = 0.0200 mol 5.000 = 4.752 + log [x / (0.02 - x)] x = 0.01278 mol <--- that's the acetate moles 0.02 - 0.01278 = 0.00722 mol acetic acid moles Determine the moles of HCl to be added: MV = (0.490 mol/L) (0.00630 L) = 0.003087 mol The HCl will protonate the Ac¯ Ac¯ ---> 0.01278 - 0.003087 = 0.009693 mol HAc ---> 0.00722 + 0.003087 = 0.010307 mol pH = 4.752 + log [0.009693 / 0.010307] pH = 4.725
Calculate the pH of a solution prepared by mixing 5.00 mL of 0.500 M NaOH and 20.0 mL of 0.500 M benzoic acid solution. (Benzoic acid is monoprotic: its ionization constant is 6.46 x 10¯5.)
Moles NaOH = (0.500 mol/L)(0.00500L) = 0.0025 Moles of benzoicA= (0.500 mol/L)(0.0200L)=0.0075 NaOH and benzoic acid react in a 1:1 molar ratio. After reaction moles of sodium benzoate = 0.00250 mol moles of benzoic acid = 0.00750 mol pH = 4.190 + log [0.00250 / 0.00750] pH = 3.713
Calculate the pH of a solution prepared by mixing 15.0 mL of 0.100 M NaOH and 30.0 mL of 0.100 M benzoic acid soluion. (Benzoic acid is monoprotic; its dissociation constant is 6.46 x 10¯5.)
Moles of NaOH = (0.100 mol/L) (0.0150 L) = 0.0015 Moles of benzoicA = (0.100 mol/L) (0.0300L)= 0.003 NaOH and benzoic acid react in a 1:1 molar ratio. After reaction: moles of sodium benzoate = 0.00150 mol moles of benzoic acid = 0.00150 mol pH = 4.190 + log [0.00150 / 0.00150]= 4.190
1.00 L of a solution containing 0.0500 mole of HOAc and 0.100 mole of NaOAc is prepared. Ignore the autoionization of water for the purposes of this problem. The Ka of HAc equals 1.77 x 10-5. (b) Which way will the reaction shift?
Since Qa < Ka, the reaction shifts right, in order to increase the value for Qa up to the value for Ka.
A buffer solution contains 5.00 mL of 2.00 M acetic acid, 45.0 mL water and 2.05 g sodium acetate. Predict the pH of the buffer solution.
The final volume of buffer is 50 mL (M1) (50 mL) = (2.0 mol/L) (5 mL) M1 = 0.2 M molarity CH3COONa (molar mass = 82.03 g/mol): 2.05/82.03 = 0.0249 mole in 0.050 L solution molarity = 0.0249 mol / 0.05 L = 0.498 M pH = 4.752 + log (0.498/0.20)=5.14
What volume of 6.00 M NaOH must be added to 0.250 L of 0.300 M HNO2 to prepare a pH = 4.00 buffer? Ka for nitrous acid: 4.0 x 10¯4
We know that this reaction will take place: HNO2 + OH¯ ---> H2O + NO2¯ (0.300 mol/L) (0.250 L) = 0.0750 mol 4.00 = 3.398 + log [x / (0.0750 - x)] x = 0.0600 moles By the 1:1 molar ratio of the overall reaction, x=moles of OH- need to be added. volume of sodium hydroxide solution required: 0.0600 mol / 6.00 mol/L = 0.0100 L = 10.0 mL
What is the pH when 25.0 mL of 0.200 M of CH3COOH has been titrated with 35.0 mL of 0.100 M NaOH?
moles of acetic acid and NaOH before mixing: CH3COOH: (0.200 mol/L) (0.0250 L) = 0.00500 mol NaOH: (0.100 mol/L) (0.0350 L) = 0.0035 mol moles of acetic acid and Naacetate+ after mixing: CH3COOH: 0.00500 - 0.00350 = 0.00150 mol CH3COONa: 0.0035 mol pH = 4.752 + log [(0.00350 mol/0.060 L) / (0.0015 mol/0.060 L)]=5.120
A buffer is prepared containing 0.800 molar acetic acid and 1.00 molar sodium acetate. What is its pH?
pH = 4.752 + log ( 1.00/ 0.800) x = 4.752 + 0.097 = 4.849
(a) Calculate the pH of a 0.500 L buffer solution composed of 0.700 M formic acid (HCOOH, Ka = 1.77 x 10¯4) and 0.500 M sodium formate (HCOONa).
pH = pKa + log [base / acid] pH = 3.752 + log [0.5 / 0.7] pH = 3.752 + (-0.146) pH = 3.606
A buffer is prepared containing 1.00 molar acetic acid and 0.800 molar sodium acetate. What is its pH?
pH = pKa + log [base / acid] x = 4.752 + log (0.800 / 1.00) x = 4.752 - 0.097 = 4.655
Fifty percent of a weak acid is in an ionized form in a solution with pH of 5.000, what is the pKa value for the weak acid?
suppose the acid is HA and its ionic anion A¯. Then, for a 50% ionized solution, pH = pKa because, at half-neutralization, [HA]/[A¯] = 1. Since log(1) = 0, the pH at half-neutralization is = pKa.