CH201-3

A generic salt, AB2, has a molar mass of 335 g/mol and a solubility of 7.60 g/L at 25 °C. What is the Ksp of this salt at 25 °C?

(7.6g/L)*(1mol/335)=.0277M AB2 --> A2+ 2B- excess 0 0 -.0227+.0277+ 2(.0277) .0277 .0454 Ksp=[A2+][B-]^2= .0277*.0454^2= .0000467

Which of the following salts has the highest solubility in water? Ignore all possible secondary acid-base reactions.

(a) Bi2S3, Ksp = 1.0 x 10-72 (b) Ag2S, Ksp = 1.6 x 10-49 (c) MnS, Ksp = 2.3 x 10-13 (d) HgS, Ksp = 1.6 x 10-54 (e) Mg(OH)2, Ksp = 8.9 x 10-12 e

In the titration of a weak acid with 0.100 M NaOH the stoichiometric point is know to occur at a pH value of approximately 10. Which of the following indicator acids would be the best choice to mark the endpoint of this titration?

(a) indicator A, Ka = 10-14 (b) indicator B, Ka = 10-11 (c) indicator C, Ka = 10-8 (d) indicator D, Ka = 10-6 b

Enough of a monoprotic acid is dissolved in water to produce a 0.0154 M solution. The pH of the resulting solution is 2.57. Calculate the Ka for the acid.

.00057 pH= -log(H+) [H+]=10^-pH 10^-2.57=.0027 M Fill in what you know HA ----> H+ + A- I .0154 0 0 C E .0027 Solve the rest HA --> H+ + A- I .0154 0 0 C-.0027+.0027+.0027 E.0127 .0027 .0027 Ka= [H+][A-]/[HA] .0027*.0027/.0127= .00057

At a certain temperature, the solubility of strontium arsenate, Sr3(AsO4)2, is 0.0540 g/L. What is the Ksp of this salt at this temperature?

.054g/L*(1mol/540.7g)= .0000999M Sr3(AsO4)2(s) 3Sr2+ 2AsO4^3- +3(9.9e-5) +2(9.9e-5) .0003 .0002 Ksp=[Sr2+]^3[AsO4^3-]^2 =.0003^3*.0002^2 =1.07e-18

Acetic acid (CH3COOH) is a weak acid (Ka = 1.8 x 10-5). Calculate the pH of a 17.6 M solution of CH3COOH (glacial acetic acid).

1.7

What is the molar solubility of AgCl? (Ksp = 1.8 × 10−10)

1.8 x 10-10 = [Ag+][Cl-] = x2; x = 1.34 x 10-5 M

If the Kb of a weak base is 1.4 × 10-6, what is the pH of a 0.16 M solution of this base?

10.68 B H2O BH+ OH- I .16 0 0 C -x +x +x E .16-x x x Kb=[BH+][OH-]/[B] =x^2/.16-x=x^2/.16 x^2=Kb(.16) x=.00047M x=[OH-]=.00047M pOH=-log[OH-]=3.32 14-3.32= 10.68

What is the pH of 8.3e-3 M HCL?

2.1 (pH = -log(H+)= -log(8.3e-3))

The flask shown here contains 0.863 g of acid and a few drops of phenolphthalein indicator dissolved in water. The buret contains 0.130 M NaOH. What is the molar weight of the acid if it took 22.5 mL of base was used to titrate.

22.5mL/1000 .0225L*.13M= .00293 mol NaOH At the second equivalence point, the added amount of OH- is equal to the total amount of H present in the acid. Since the acid is diprotic, there is 2 mol of H per mole of acid. .00293 mol NaOh* 1mol acid/2mol H+ =.00146 mol acid molar mass= .863/.00146= 5.9e2 g/mol

If Ka of HBrO is 2.8e-9, what is the value of Kb for its conjugate base?

3.6e-6 Kw=Ka * Kb Kw=1e-14 Kb=1e-14/2.8e-9= 3.6e-6

The flask shown here contains 0.922 g of acid and a few drops of phenolphthalein indicator dissolved in water. The buret contains 0.150 M NaOH. If the titration takes 32mL base, what is the molar mass of the acid?

32 mL/1000*.15M= .0048 mol Since the acid is monoprotic, number of mols of NaOH is equal to the mol of Acid in the flask. .9022g/.0048mol=192g/mol

If a buffer solution is 0.290 M in a weak acid (Ka = 4.1 × 10-5) and 0.410 M in its conjugate base, what is the pH?

4.54 Ka=[H+][A-]/[HA] 4.1e-5=[H+]*.42/.29 [H+]=.000029 pH=-log[H+] =4.54 or pKa=-log(Ka) =-log(4.1e-5)=4.39 pH=pKa+log[A-]/[HA] =4.39+log[.41]/[.29] =4.54

A 50.0 mL sample of 0.500 M HCN (Ka = 6.2 x 10-10) is titrated with 0.500 M NaOH. What volume of 0.500 M NaOH is required to reach the stoichiometric point?

50.0mL

A 500.0 mL sample of 0.500 M HCN (Ka = 6.2 x 10-10) is titrated with 0.500 M NaOH. What volume of 0.500 M NaOH is required to reach the stoichiometric point?

500mL

A 5.95 g sample of an acid (H2X), requires 45.0 mL of a 0.500 M NaOH solution for complete reaction (the removal of both protons). The molar mass of the acid is:

529g/mol

What is the pH of 8e-8 M HCL?

6.8 (you have to take into consideration ionization of water and the shift of equilibrium) 8e-8+1e-7= 1.8e-7 pH = -log(1.8e-7)=6.7447 take into consideration h20 ---> H+ + OH- +x -x -x This reduces the H+ in solution, raising pH to 6.8 roughly.

A 75 mL sample of 0.0500M HCN (Ka = 6.2 x 10-10) is titrated with 0.500 M NaOH. What volume of 0.500 M NaOH is required to reach the stoichiometric point?

7.5 mL

find H3O+ if OH- is 1.25e-7 M

8e-8 (1.0e-14/1.25e-7)

For the following stoichiometric titrations in aqueous solution, identify whether the final pH will be >, <, or = 7.0. (a) KOH (aq) + CH3COOH (aq) (b) C6H5NH2 (aq) + HCl (aq) (c) HBr (aq) + LiOH (aq)

> 7 < 7 = 7

Calculate standard free energy change 2Au3+ + 3Zn ↔ 2Au + 3Zn2+

Anode 3Zn → 3Zn2+ + 6e- Cathode 2Au3+ + 6e- → 2Au E°cell= 1.498+.76= 2.26V delG°= -nFE°cell =-(6)(96485)(2.26)= -1310000J

Ksp of A_3B_2(s)

A3B2 3A2+ 2B3- excess 0 0 -x +3x +2x 3x 2x Ksp=[A2+]^3[B3-]^2 3x^3*2x^2=108x^5

At 25 °C only 0.0160 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C?

AB3 -> A3+ 3B- 0 0 -.016 +.016 +3(.016) .016 .048 Ksp=.016*(.048^3) =1.77e-6

What concentration of SO32- is in equilibrium with Ag2SO3(s) and 2.50 × 10-3 M Ag ? The Ksp of Ag2SO3 can be found here.

Ag2SO3(s)-> 2Ag+ SO3^2+ the solubility expression is Ksp=[Ag]^2[SO3^2-] [SO3&2-]=Ksp/ [Ag+]^2 The solubility constant of silver(I) sulfite is 1.50 × 10-14. Solving the solubility expression for [SO32-] and substituting values gives =1.5e-14/2.5e-3= 2.4e-9M

A solution is 5.0× 10-5 M in each of these ions: Ag , SO42-, Cl-, CO32- Which precipitate will form?

Ag2SO4 (Ksp = 1.12× 10-5) AgCl (Ksp = 1.77× 10-10) Ag2CO3 (Ksp = 8.46× 10-12) AgCl

Calcium sulfate is slowly added to a solution containing 0.0500 M Ca2 (aq) and 0.0320 M Ag (aq). What will be the concentration of Ca2 (aq) when Ag2SO4(s) begins to precipitate? Solubility-product constants, Ksp, can be found here.

Ag2SO4 ↔ 2Ag+ SO4^2- Ksp=[Ag+]^2[SO4^2-]=1.2e-5 [SO4^2-]=1.2e-5/.032^2= .0117M CaSO4↔ Ca2+ SO4^2- Ksp=[Ca2+][SO4^2-]=4.93e-5 [Ca2+]=4.93e-5/.0117=.00421 % precipitated= .05M-.00421M/.05M*100%= 91.6%

HCl is slowly added to a solution containing 0.01 M Pb2+ and 0.01 M Ag+. Given Ksp (PbCl2) = 2.4 x 10-4 and Ksp (AgCl) = 1.6 x 10-10, which compound precipitates first?

AgCl

If you have a Anode of Pb submerged in KCL and a cathode of Ag submerged in KCL with precipitate in both solutions of PbCl2 and AgCl respectively, what is anode and cathode half reaction and shorthand notation for the net cell reaction.

Anode Pb + 2Cl- ↔ PbCl2 + 2e- Cathode 2AgCl + 2e- ↔ 2Ag + 2Cl- Shorthand Pb[PbCl2[Cl-[]Cl-[AgCl[Ag

Sn[Sn2+[]Ag+[Ag Anode and Cathode half reactions and net cell reaction

Anode Sn→Sn2+ + 2e- Cathode Ag+ + e- → Ag Net 2Ag+ + Sn ↔ 2Ag + Sn2+

Which of the following salts has the lowest solubility in water? Ignore all possible secondary acid-base reactions.

Bi2S3, Ksp = 1.0 x 10-72

The pKa's for CH3COOH and HNO2 at 25°C are 4.74 and 3.35, respectively. Which acid produces the stronger conjugate base?

CH3COOH

Current is applied to an aqueous solution of calcium bromide. What is produced at the cathode and the anode?

Ca2+ can be reduced Br- can be oxidized H2O can go either way Reduction potentials yield Ca2+ → Ca = -2.87 V 2H20→ H2 + 2OH- =-.83 2Br- → Br2 = 1.07 2H20→ O2 +4H+ = 1.23 The reduction of water is favored because it has a higher potential. H+ is produced at the cathode. The oxidation of bromide is favored because it has a lower potential. Br2 is produced at the anode.

For which of the following mixtures will Ag2SO4(s) precipitate? 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.20 M AgNO3(aq) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.30 M AgNO3(aq) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.40 M AgNO3(aq) 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.50 M AgNO3(aq)

Calculate Qsp for each mixture. Precipitation occurs if Qsp > Ksp. In each case, the total volume is 155.0 mL, and the SO42- concentration immediately after mixing is .097MSO4^2- Now find Qsp for each choice and compare it to Ksp. Note that Ksp = 1.2× 10-5. Use molararity to find Ag in Eq and solve Qsp Qsp=[Ag+]^2[SO4^2-] if this is larger than Ksp, precipitate forms.

Co[Co2+(.0155M)[]Ag+(3.5M)[Ag Write the net cell equation Find E°cell, Ecell, del G° of reaction, and del G of reaction

Co + 2Ag+↔2Ag + Co2+ E° of Co=-.28 V E° of Ag=.80V E°cell=.8--.28=1.08 V del G°= -nFE°cell= -(2)(96485C/mol)(1.08V)=208000J=-208KJ/mol del G= del G°+RTln(Q) = -208+(8.3145e-3KJ/mol)(298)*ln(.0155/3.5^2)=-225KJ/mol E cell= -del G/nF= -225000J/(2*96485)=1.17 V

A current of 5.23 A is passed through a Cr(NO3)2 solution. How long (in hours) would this current have to be applied to plate out 7.50 g of chromium?

Cr2+ + 2e- → Cr 7.5g*(1molCr/51.996g)*(96485C/mol of e-)*(2 mol e-/mol Cr)= 27800 C 27800/5.23/3600sec= 1.48 hour

Calculate E°cell Cu + Ag+ → Cu+ + Ag

Cu+[Cu = +.52V Ag+[Ag = +.80V E°cell = E°cathode-E°anode .80-.52=.28V

Copper(I) ions in aqueous solution react with NH3(aq) according to Cu+ 2NH3 ↔ Cu(NH3)2 + Kf=6.3e10 Calculate the solubility (in g·L-1) of CuBr(s) (Ksp = 6.3× 10-9) in 0.28 M NH3(aq).

CuBr ↔ Cu+ CuBr- Ksp=6.3e-9 Cu+ 2NH3 ↔ Cu(NH3)2 + Kf=6.3e10 CuBr 2NH3 ↔ Cu(NH3)2 + Br- K=? K=Ksp*Kf=397 CuBr(s) 2NH3 ↔ Cu(NH3)2 Br- .28 0 0 -2x +x +x .28-2x x x K=[Cu(NH3)2^+][Br-]/[NH3]^2 397=x^2/(.28-2x)^2 x=.14M .14mol/L*(143.45g/mol)=20g/L

Ksp of CuCO3 is 1.4e-10 What is its solubility in g/L

CuCO3>Cu2+ CO3^2- 0 0 -x +x +x x x Ksp=[Cu2+][CO3^2-] =x^2 x=sqrt(1.4e-10) x=1.2e-5M 1.2e-5mol/L*(123.554g/mol)= .0015g/L

Determine the potential of the following galvanic cell under standard conditions: Zn(s) | Zn2+(1.00 M) || Cu2+(1.00 M) | Cu(s)

E0cell = E0cathode - E0anode = +0.342 V - (-0.762 V) = +1.10 V

Zn[Zn2+(.1M)[]Zn2+(xM)[Zn Voltage equals 12 mV at 298K. calculate concentration of Zn2+

Ecell=E°cell-RT/nF*lnQ 12e-3=0-(8.3145*298)/(2*96485)*ln(.1/x) 12e-3/-.01284=ln(.1/x) e^-.935=.1/x x=.255M

What is the Ecell of Zn + Sn2+ ↔ Zn2+ + Sn [Zn2+]=.876M [Sn2+]=.018M

E°cell=-.14+.76= .62V Ecell=E°-RT/nF*ln(Q) Q=[Zn2+]/[Sn2+]=48.7 E=.62V-(8.314*298)/ (2*96485)*ln(48.7) = .57 V

Concentration cells operate on the principle that standard reduction potentials are dependent on concentration.

False

In an electrolytic electrochemical cell, chemical energy is converted into electrical energy

False

Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. Calculate the pH for each of the following points in the titration of 50.0 mL of a 3.0 M H3PO3(aq) with 3.0 M KOH(aq). pKa1=1.3 pKa2=6.7 before addition of KOH after 25mL KOH after 50mL KOH after 75mL KOH after 100mL KOH

From the given values of pKa1 and pKa2 , we find that the values of Ka1 and Ka2 for phosphorous acid are 10-1.30 = 0.050 and 10-6.70 = 2.0× 10-7, respectively. (a) Because Ka1 >> Ka2 , we can ignore the contribution of the second ionization to the value of [H ]. Since Ka1 is so much larger, we ignore Ka2 H3PO4-> H+ H2PO3- I 3.0 0 0 C -x +x +x E 3-x x x Ka1=[H+][H2PO3-]/[H3PO3] =x^2/3-x quad exp yields x=.36 so the pH is .44 After 25mL of 3M KOH the number of mols is half the number of mols of H3PO3, so this is the first midpoint. thus, pH=pKa1=1.30 after 50mL of 3M KOH, the number of mols of KOH is equal to the number of moles of H3PO3, so this is the first EQ point. Thus, pH=pKa1+pKa2/2=4 after 75 mL this is the second midpoint, thus pH=pKa2=6.70 After 100mL, mols of OH- is twice mols H3PO3, second EQ point.

The Kc for the following reaction is 7.0 x 105 at 600 K: H2(g) + Cl2(g) ! 2 HCl(g) Starting with 5 moles of HCl(g) in a 5-L reaction vessel at 600 K, calculate the equilibrium concentrations of H2, Cl2, and HCl.

H2(g) + Cl2(g) ! 2 HCl(g) I 0M 0M 1M C +x +x -2x E x x 1-2x K = 7.0 x 105 = (1-2x)2 / x2 √7.0 x 105 = √(1.0 - 2x)2 / x2 836.7 x = 1.0 - 2x 838.7 x = 1.0 x = 1.19 x 10-3 M = [H2] = [Cl2] [HCl] = 1.0 - 2x ~ 1.0 M

For the diprotic weak acid H2A, Ka1 = 2.4 × 10-6 and Ka2 = 5.4 × 10-9. What is the pH of a 0.0500 M solution of H2A? What are the equilibrium concentrations of H2A and A2- in this solution?

H2A-->H+ + HA- .05 0 0 -x +x +x .05-x x x Ka=[H+][HA-]/[H2A] =x^2/.05-x x is approximately sqrt(2.4e-6*.05)= .00035M pH=-log(.00035)=3.46 HA- --> H+ + A2- x x 0 -y +y +y x-y x+y y Ka2= [H+][A2-]/[HA-] (x+y)(y)/(x-y)=y y=Ka2=5.4e-9 M [H+]=x+y=.00035+5.4e-9 =.00035 [H2A]=.05M - x =.05-.00035=.0497M [A2-]=y=5.4e-9M

The conjugate acid and conjugate base of the bicarbonate ion, HCO3-, are respectively:

H2CO3 and CO3^2-

What are the conjugate acid and base of HPO42-?

H2PO4- (c. acid) and PO43- (c. base)

In the titration of a weak acid with 0.100 M NaOH the stoichiometric point is known to occur at a pH value of approximately 7. Which of the following indicator acids would be the best choice to mark the endpoint of this titration?

Indicator C, Ka = 10-8

In the titration of a weak base with 0.100 M HCl the stoichiometric point is known to occur at a pH value of approximately 7. Which of the following indicator acids would be the best choice to mark the endpoint of this titration

Indicator D, Ka = 10-6

Phosphoric acid, H3PO4(aq), is a triprotic acid, meaning that one molecule of the acid has three acidic protons. Estimate the pH, and the concentrations of all species in a 0.350 M phosphoric acid solution. pKA1= 2.16 pKA2= 7.21 pKA3= 12.32

KA1= 6.9e-3 KA2= 6.2e-8 KA3= 4.8e-13 KA2 and KA3 are so small, only use KA1 for [H+] H3PO4 -> H+ H2PO4 I .350 0 0 C -x +x +x E .35-x x x KA1= x^2/.35-x 6.9e-3 x= .046 = [H+] pH= -log(.046)=1.34 [OH-]=1e-14/.046 =2.2e-13 [H3PO4]=(.35-.046)M =.304M H2PO4- ->H+HPO42- I .046 .046 0 C -y +y +y E .046-y .046-y y KA2= y(.046+y)/(.046-y) =6.2e-8 y=6.2e-8 [HPO4^2-]=6.2e-8M [H2PO4^-]=.046-6.2e-8=.046M HPO4^2- >H+PO4^3- I 6.2e-8 .046 0 C -z +z +z E 6.2e-8-z .046+z z KA3= z(.0460/ (6.2e-8)-z=4.8e-13 z=6.5e-19M= [PO4^3-]

If the Kb for NH3 is 1.76 x 10-5, calculate the pKa for NH4+.

Ka = Kw/Kb = 5.68 x 10-10; pKa = -log(5.68 x 10-10) = 9.25

What is the molar solubility of Mg(OH)2 at pH 11.0?

Ksp = 1.8 x 10-11 Mg(OH)2(s) ↔ Mg2+(aq) + 2 OH-(aq) pH = 11.0; pOH = 3.0; [OH-] = 1 x 10-3 M Ksp = [Mg2+][OH-]2 = 1.8 x 10-11 Mg(OH)2(s)↔ Mg2+(aq) + 2 OH-(aq) 0 1.0e-3 +x +2x x 1.0e-3+2x Ksp = 1.8 x 10-11 = (x)(1x10-3)2 x = 1.8 x10-11 / (1x10-3)2 = 1.8 x 10-5 M, which is much lower than the true molar solubility of Mg(OH)2 at neutral pH (1.65 x 10-4 M).

What is the molar solubility of PbCl2, Ksp = 1.60 × 10−5?

Ksp = [Pb2+][Cl-]2 = (x)(2x)2 = 4x3 = 1.6 x 10-5 x = 3√(1.6 x 10-5/4) = 0.0175 M

If the Ka for HCN is 6.2 x 10-10, calculate Kb for CN-.

Kw = KaKb Kb = Kw/Ka = 1 x 10-14 / 6.2 x 10-10 = 1.6 x 10-5

What is the pH of 0.50 M NH4Cl? (Kb (NH3) = 1.76 x 10-5)?

NH4+ + H2O → NH3 + H+ Ka = Kw/Kb = 1 x 10-14 / 1.76 x 10-5 = 5.68 x 10-10 Ka = 5.68 x 10-10 = [NH3][H+]/[NH4+] = x2 / 0.50 x2 = (5.68 x 10-10)(0.50) x = 1.68 x 10-5 M = [H+] pH = -log(1.68 x 10-5 M) = 4.77

NH3 is a weak base (Kb = 1.8 × 10-5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.077 M in NH4Cl at 25 °C?

NH4Cl dissociates into the acidic NH4 ion and the neutral Cl- ion. Since NH3 and NH4 are conjugates, their K values are related. Ka*Kb=Kw, so Ka= 5.556e-10 HA↔ H+ A- I .077 0 0 C -x +x +x E .077-x x x Ka=[H+][A-]/[HA] =x^2/.077-x =x^2/.077= 5.56e-10 x=6.5e-6M pH=-log[6.5e-6] =5.18

A 0.170-mole quantity of NiCl2 is added to a liter of 1.20 M NH3 solution. What is the concentration of Ni2 ions at equilibrium? Assume the formation constant* of Ni(NH3)62 is 5.5 × 10^8.

Ni2+ 6NH3↔ Ni(NH3)6^2+ all aq .17 1.2 0 -.17 -6(.17) +.17 x .18 .17 Kf=[Ni(NH3)6^2+]/ [Ni2+][NH3]^6 5.5e-8=.17/.18^6*x x=9.1e-6 M Ni2+

Which of the following will be more soluble in an acidic solution than in pure water?

Of the anions listed, hydroxide, sulfite, and sulfate can react with H ions. Perchlorate and iodide cannot react with H because they are conjuages of strong acids. In the solubility equilibrium shown here, a shift toward the right leads to increased solubility. If the anion reacts, it\'s concentration decreases, and the equilibrium shifts right according to Le Châtelier\'s principle. An acidic solution has an abundance of H ions, which can react with basic anions such as hydroxide, sulfite, and sulfate. BaSO4 Ni(OH)2 BaSO3

Ksp of PbBr2 is 6.6e-6 What is its molar solubility in pure water? What is its solubility in .5M KBr solution? What is the molar solubility in .5M Pb(NO3)2 solution?

PbBr2 ↔ Pb2+ 2Br- ex 0 0 -x +x +2x x 2x Ksp=[Pb2+][Br-]^2 6.6e-6=[x][2x]^2=4x^3 x=1.18e-2M In .5M KBr PbBr2 Pb2+ 2Br- ex 0 .5 -x +x +2x x .5+2x Ksp=[Pb2+][Br-]^2 6.6e-6=(x)(.5+2x)^2 assume .5+2x=.5 x=6.6e-6/.5^2 x=.0000264M In .5M Pb(NO3)2 PbBr2 ↔ Pb2+ 2Br- ex .5 0 -x +x +2x .5+x 2x Ksp=[Pb2+][Br-]^2 6.6e-6=(.5+x)(2x)^2 assume x is small again 6.6e-6=2x^2 x=.00182

Methyl orange is an indicator with a Ka of 1 x 10-4. In its acid form (HIn) the compound is red whereas in its base form (In-) the compound is yellow. At pH = 3.0, this indicator will be:

RED

Current is applied to a molten mixture of PbF2, FeCl2m and AlBr3 What forms at the cathode and anode?

The cations are Pb2+, Fe2+, and Al3+. The reduction potential of Pb2+ is the highest, so it is favored at the cathode. Reduction produces Pb The anions are F-,Cl-,and Br-. The reduction potential for Br- is the lowest, so its oxidation potential is the highest. oxidation of Br- produces Br2 at the anode

At a concentration of 1 M, the weak acid HNO2 is 2% ionized and the pH of the solution is 1.7. What happens when KNO2(s) is dissolved into the solution?

The extent of HNO2 ionization decreases, Concentration of H+ decreases and thus the pH increases. HNO2↔H+ + NO2^- Adding KNO2 adds NO2^- to the solution to shift it to right.

When aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a solid precipitate forms. However, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble [Pb(OH)4]2-(aq) complex ion.

The first tricky part about this question is determining the formula for the ionic compounds. Sodium forms 1 ions, Na , and the Roman numeral in lead(II) indicates a 2 charge, Pb2 . The polyatomic ions nitrate and hydroxide both have a -1 charge, NO3- and OH-. Thus, sodium hydroxide is NaOH, lead nitrate is Pb(NO3)2, sodium nitrate is NaNO3, and lead hydroxide is Pb(OH)2. Since all nitrates are soluble, the precipitate must be lead hydroxide. For the second equation, solid lead hydroxide reacts with hydroxide (putting either OH- or NaOH was acceptable in this case) to form the complex ion. 2NaOH+Pb(NO3)2→Pb(OH)2+2NaNO3 Pb(OH)2+2OH-→[Pb(OH4]^2-

CaCO3 is more soluble in 1.0 M H+ than it is in pure water.

True

The pOH of a 0.10 M solution of Ba(OH)2 is 0.70.

True

When is a cell reaction spontaneous?

When Voltage is positive

Above what Fe2 concentration will Fe(OH)2 precipitate from a buffer solution that has a pH of 9.69? The Ksp of Fe(OH)2 is 4.87×10-17.

When iron(II) hydroxide dissolves in water, it forms Fe2 cations and OH- anions. Fe(OH)2 ↔ Fe2+ 2OH- Ksp=[Fe2+][OH-]^2 pOH=4.31 [OH-]=.000049M Ksp=[Fe2+][OH-]^2 4.87e-17=x*.000049^2 x=2.0e-8M

At 22 °C an excess amount of a generic metal hydroxide M(OH)2 is mixed with pure water. The resulting equilibrium solution has a pH of 10.28. What is the Ksp of the salt at 22 °C?

When the metal hydroxide dissolves in water, it forms M2 cations and OH- anions. The pOH can be determined from the pH. pOH=3.72=-log[OH-] [OH-]=.00019M According to the balanced chemical reaction, the M2 concentration is half the OH- concentration. .000095mol/L M2+ [M2+]=.000095M Ksp=[M2+][OH-]^2= .000095*.00019^2 Ksp=3.5e-12

Methyl orange is an indicator with a Ka of 1 x 10-4. In its acid form (HIn) the compound is red whereas in its base form (In-) the compound is yellow. At pH = 6.0, this indicator will be:

Yellow

In the following redox equation, which species is the reducing agent? Zn(s) + Cd2+(aq) ↔ Zn2+(aq) + Cd(s)

Zn(s)

What does the pH equal at the equivalence point?

corresponds to a solution of the conjugate base in water. (A-)

What does the pH equal at the initial point in titration?

corresponds to a solution of the weak acid in water (HA)

Calculate the approximate equilibrium constant at 25°C for the following reaction 2 Al(s) + 3 Cu2+(aq) → 2 Al3+(aq) + 3 Cu(s)

log K = nE0rnx/0.059 = (6)(2.0) / 0.059 = 203 K ~ 10^203

Gallium is produced by the electrolysis of a solution obtained by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(III) solution by a current of 0.800 A that flows for 90.0 min.

m=(current*time/Faraday's)*(molar mass/#of electrons transferred) m=(It/F)*(mew/n) Ga3+ + 3e- → Ga n=3 m=1.04 g

An aqueous solution has a proton ion concentration of 5.0 x 10-6 M, what is its pH?

pH = -log(5.0 x 10-6) = 5.3

You have 100 mL of a 0.1 M solution of an unknown weak acid HA. After adding 30 mL of 0.10 M NaOH, the pH of the solution is 6.50. What is the Ka value of this unknown acid HA? This can be done 2 ways, as below or using the same numbers at equilibrium plugged into the HH EQ

pH = 6.50 = - log [H+]; [H+] = 3.2 x 10-7 M mol HA = (0.1 mol/L)(0.100 L) = 0.01 mol mol OH- added: (0.1 mol/L)(0.030 L) = 0.003 mol Total volume after addition = 0.130 L HA + OH- ! A- + H2O I 0.01 mol 0.003 mol 0 mol C -0.003 mol -0.003 mol +0.003 mol E 0.007 mol 0 mol 0.003 mol We know [HA] = 0.007 mol/0.130 L = 0.054 M [H+] = 3.2 x 10-7 M and [A-] = 0.003 mol/0.130 L = 0.023 M and that: HA ! H+ + A- Therefore: Ka = [H+][A-]/[HA] = (3.2 x 10-7)(0.023)/0.054 = 1.36 x 10-7

What is the proton ion concentration of an aqueous solution with a pH of 9.5?

pH = 9.5 = -log[H+]; [H+] = 3.2 x 10-10 M

What is the pH of a 0.2 M HF/0.4 M NaF buffer solution (Ka of HF = 7.2 x 10-4)?

pH = pKa + log([base]/[acid]) = 3.14 + log(0.4/0.2) = 3.14 + 0.3 = 3.44

What is the pH of a 0.5 M HF/0.5 M NaF buffer solution (Ka of HF = 7.2 x 10-4)?

pH = pKa + log(base/acid) = -log(7.2 x 10-4) + log (1) = 3.14

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.180 M HClO(aq) with 0.180 M KOH(aq). The ionization constant for HClO can be found here. (a) before addition of any KOH (b) after addition of 25.0 mL of KOH (c) after addition of 30.0 mL of KOH (d) after addition of 50.0 mL of KOH (e) after addition of 60.0 mL of KOH

pH is determined by the dissociation of HCIO HCIO -> H+ CIO- I .18 0 0 C -x +x +x E .18-x x x Ka=[CIO-][H+]/[HCIO]= x^2/.18M-x=x^2/.18 x=[H+]=.000085 pH=4.07 with equal concentrations of monoprotic titrant and analyte, the EQ point would occur when the volumes are equal. 50mL KOH at 25mL, this is the first midpoint. thus, pH=pKa=-log(4e-8)=7.4 At 30 mL we are at 30/50 concentration so 60 percent. this means that 40 percent of the acid still remains and 60 percent of the conjugate base has formed. thus the pH=7.4+log(60/40)=7.57 At 50 mL, we are at the EQ point. all of the acid has reacted. thus the pH is determined by the base. 50mL*.18M=9mmol CIO- [CIO-]=9mmol/100mL=.09 M Kb=1e-14/4e-8=2.5e-7 [OH-]=sqrt(.09*2.5e-7)=3.82 pH=10.18 At 60 mL is beyond EQ, excess KOH. we can ignore CIO- because KOH is a strong base. 50 mL reacted with the acid, so only 10mL is used in the calculations. m1v1=m2v2 m2=.18M*10ml/60ml+50ml .0164 M OH- pOH=1.786 pH=12.21

Calculate the change in pH when 9.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). A list of ionization constants can be found here.

pH=Pka+log[b/a] deltapH=pKa+ log[b]/[a] .1M NH3*100ml= 10mmol base .1M NH4+*100ml= 10 mmol acid delta pH= log(10-.9)mmol/ (10+.9) mmol log(9.1)/(10.9)= -.08 other way around makes .08

What does the pH equal at the first midpoint?

pH=pKa

You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.140 M sodium benzoate. How much of each solution should be mixed to prepare this buffer?

pH=pKa+log(base/acid) 4.00=4.2+log(benzoate/benzoic acid) log(b/Ba)=4-4.2 10^-.2=.63 x=volume x(.14M)/(.1L-x)(.1M) =.63 x=31 mL benzoate 100 ml-31= 69mL benzoic acid

A 25.00 mL sample of 0.280 M LiOH analyte was titrated with 0.750 M HCl at 25 °C. Calculate the initial pH before any titrant was added.

pOH=-log(.28)=.553 pH=13.45 25mL*.75M=7mmol 5mL*.75M=3.75mmol 7-3.75=3.25mmol after the reaction occurs M=3.25/25+5mL= .108M LiOH pOH= -log(.108)= .965 pH=13.03

As water is heated, its pH decreases. Therefore, the dissociation of water is an endothermic process.

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A fuel cell designed to react grain alcohol with oxygen has the following net reaction with 12 electrons being transferred overall: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) The maximum work one mole of C2H5OH(l) can yield by this process is 1320 kJ. What is the theoretical maximum voltage this cell can achieve?

ΔGºcell = -nFEocell = -1320 kJ/mol = -(12)(96500 C/mol) (E°cell) E°cell = (1320000 CV/mol) / (12)(96500 C/mol) = 1.14 V