Acids and Bases (B)

Consider the titration curve below for the titration of a 25.0 mL sample of trimethylamine, (CH3)3N, with 0.100 mol L−1HCl(aq). Trimethylamine is a weak base with Kb= 6.3×10−5. (a) What is the neutralization reaction? (b) Describe how the composition of the solution changes as the titration proceeds. (c) What is the concentration of the trimethylamine solution?

(CH3)3N + HCl --> (CH3)3NH + Cl "CH3 = Methyl = Me" Starting comp: Me3N = Co Half comp: >> Buffer solution. >> 1:1 for conj pair equal conc's. >> Not half of Co because of the change in volume >>> [Co x vi ]/ 2 x (vi+15) ^^ The 2 below is the half; vi = 25 >> From stoich, [Me3NH}eq = [Cl-] Eq comp:[Me3NH]eq = [Cl-] = Co x 25 / (25+30) Excess comp: >> Mostly determined by HCl which ionizes into H+ and Cl- >>[Cl-]eq = [Me3NH]eq + [H3O+]

Without doing any pH calculations, arrange the following solutions in order of increasing pH (i.e., from lowest to highest pH). Assume that each solution is prepared by dissolving 1.0 molof the salt in water to give 1.0 L of solutions at 25 °C K2S(aq)NH4Cl(aq)NH4NO2(aq)KNO2(aq)

1. 2K+ + S2- >> Basic >> Kb = Kw/Ka2 >>1 2. NH4+ + Cl- >> Cl- con B of HCl so very weak >> Acidic ( Kw/Kb) 3. NH4+ + NO2 >> NH4+ is weak acid from last one >> NO2 base (Kb=Kw/Ka) >> Base eq stronger 2 < 3 < 4 < 1

With the assumption that HBr is 100% ionized in aqueous solution, what is the pH off 1e-7molL HBr (aq) at 25oC? A. 7.00 B. 6.70 C. < 6.7 D. >6.7 -- Presss i for data

1e-7 HBr = 1e-7 H3O+ and Br- But Kw also in effect, which is a source of H3O+, so need to add that But this H2O was there initially (Kw), so adding the H3O+ by adding the strong acid moves equilibrium left! This decreases [OH-] Since [OH-] is only from water, water's contribution decreases, and it <1e-7 so from Hbr >> 1e-7 + <1e-7 So it will be less than 2(1e-7)

What is a buffer solution? How?

A solution that contains a weak conjugate acid base pair Resist changes in pH when strong species added Adding strong will cause a neutralization reaction -- HA + NaA HA + NaOH BHCl + B HCl + B

Starting with 0.1 mol/L HNO2(aq), which one of the following operations would cause the pH of the solution to increase? A.adding some NaNO2 B.adding some HCl C.adding some NaCl D.adding some HNO -- Ka for HNO2is 7.2x10−4

A. Add a weak base (NO2) = pH goes up B. Decrease pH by increasing H3O+ C. Na and Cl do nothing D. pH decreases *A*

When is it appropriate to set up and use the following equilibrium to calculate the pH of the solution? CH3COOH(aq) + H2O(l) ⇌CH3COO−(aq) + H3O+(aq) A. When dissolving CH3COOH in water B. Whendissolving CH3COOH and NaCH3COO in water C. When dissolving CH3COOH and a smaller amount of NaOH in water D. Before the equivalence point in a titration of CH3COOH with NaOHE.all of the abov

A. YES B. YES C. YES, first do neturalization reaction to make salt then do salt's one, same as B D. Same as B *E.*

When is it appropriate to set up and use the following equilibrium to calculate the pH of the solution? Check all that apply.CH3COO−(aq) + H2O(l) ⇌CH3COOH(aq) + OH−(aq) A.when dissolving CH3COONain water B.when dissolving CH3COOH and CH3COONa in water C.when equal moles of CH3COOH and NaOH are dissolved in water (e.g. at the equivalence point in the titration of CH3COOH with NaOH) D.when dissolving CH3COOH and excess NaOHin water (e.g. past the equivalence point in the titration of CH3COOH with NaOH

A. Yes, pH via pOH B. Yes, this is a buffer solution. Same as (a), but with another step >> Doesnt matter which side of the same equilibrium you use >> Can solve for Kb C. Yes, this is equivalence point >> Neutralization rxn that goes 100% to make CH3COOH- + Na+ + H2O >> Same as (a) D. No Excess NaOH past equivalence point, its the NaOH is main determinant and CH3COOH is used up

When NH3(aq) is titrated with HNO3(aq) at 25 oC, the pH at the equivalence point is A. less than 7 B. equal to 7 C.greater than 7 When HCl(aq) is titrated with KOH(aq) at 25 oC, the pH at the equivalence point is A. less than 7 B. equal to 7 C.greater than 7

After neutralization : NH3 + HNO2 --> NH4 + NO3- >> NH4 weak acid = but still determines more than NO3- since NO3- is super weak >> NO3- weak >> Less than 7 -- After neutralization: HCl + KOH --> Cl- + K+ + H2O Neutral = 7 If temp went higher, increased [H+] so lower pH

What is a buffer capacity (BC)? [BC] does what as [buffer] increases What is a buffer range?

Amount of acid or base that a buffer can neturalize before pH changes appreciably [BC] increases as [buffer] increases pH range over which a buffer maintains a fairly constant pH >> pH = pKa +/-1

Titration of 50mLof 0.05M of CH3COOH (Ka - 1.8e-5) with 0.1M NaOH. Calculate the pH of solution at: a) 0mL NaOH b) 5mL NaOH c) 10mL NaOH d) At equivalence point e) 10mL past the equivalence point

CH3COOH + H2O >< CH3COO- + H3O+ *a)* i. 0.05 | na | 0 | 0 c. -x | +x | +x e. 0.05 | x | x >> SSA Ka = x^2 / 0.050 x = sqr (Kax0.050) = 9.5e-4 = [H3O+] pH = 3.02 *b)* nNaOH = 0.1M x 5mL = 0.5mmol Vtotal = 55mL nCH3COOH = 50mL x 0.050mol/L = 2.5mmol CH3COOH + NaOH --> CH3COONa + H2O mix. 2,5mmol | 0.5mmol get. 2.0mmol | na | --> 0.5mmol >> this is a buffer! CH3COOH + H2O >< CH3COO- + H3O+ i. 2.0/55 | na | 0.5/55 | na c. -x | +x | +x e. 2.0/55 | na | 0.5/55 | x >> SSA >> concentrations are same b/c mainly determined by neutralization Ka = {[0.5/55]x} / (2.0/55) >> 55's cancel x = 7.25e-5 = [H3O+] pH = 4.14 -- could also jusmp to pH = (-logKa) + log(0.5/2.0) >> Since conjugate base pair from neutralization *c)* Vnaoh = 10mL ; vtotal = 60mL molNaOH = 10mL x 0.1M = 1.0mmol CH3COOH + NaOH --> CH3COONa + H2O mix: 2.5mmol | 1mmol --> na | na get: 1.5mmol | 0mmol --> 1mmol | na >> Just to H-H equation since using mmol and volumes cancel pH = pKa + log [A-] / [HA] pH = pKa + log [1.0][/[1.5] pH = 4.57 >> or do ice i. 1.5/60mL | na --> 1.0/60mL | na c. -x | na --> +x | +x e. 1.5/60 | na --> 1.0/60mL | x >> SSA Ka = [(1.5/60) x] / (1.5/60) x = 2.e-5 = [H3O+] Molar ratio determined by mole ratio *d)* Equivalence is point nNaOH = nCH3COOH >> where nCH3COOH = 50mL x 0.05M = 2.5mmol >> so we need 2,5mmol Vnaoh Vnaoh = 2.5mmol x (L/0.1mol) = 25mL so Vtotal =75mL (50mL+25mL) mix. 2.5mmol | 2,5mmol --> 0 | na get. 0 | 0 --> 2.5mmol na We make a weak base CH3COO- + H2O >< CH3COOH + HO- i. 2.5mmol/75 | na --> 0 | na c. -x | na --> +x | +x e. 2.5/75 | x | x >> SSA ^^ KaxKb=Kw, find Kb which is 10e-9 Kb = x^2 / (2.5/75) Kw/Ka = x^2 / (2.5/75) x = sqr(Kw/Ka x (2.5/75)) = 4.3e-6M x = HO- pOH = 5.37 pH + pOH = pKw 14-5.37 = pH = 8.63 *e)* Vnaoh = 35mL Vtotal = 85mL (from 50+35) nNaOH = 35mL x 0.1M = 3.5mmol CH3COOH + NaOH --> CH3COONa + H2O mix. 2.5mmol | 3.5mmol --> 0 | na get. 0 | 1.5mmol --> 2.5mmol |na Strong base weak base, strong base more important (always go with species with highest equilibrium constant) NaOH --> NA+ + HO- i. 1mmol/85 --> na | 0 f. 0 --> 1/85 | 1/85 >> 1:1 ratio [HO-] = 0.012M pOH = 1.92 pH = 12.08

Calculate % ionization of CH3COOH in a solution prepared by dissolving: a) 0.0125mol of CH3COOH or b) 0.0125 mol CH3COOH and 0.0025 mol CH3COONa in water to make a 1L solution Ka (CH3COOH) = 1.8e-5

CH3COOH + H2O >< CH3COO- + H3O+ A) I) 0.0125 | na | 0 | 0 C) -x | +x | +x E) 0.0125-x | x | x >> SSA (from Ka=e-5) for 0.0125-x=0.0125 Ka = x^2 / 0.0125 x = sqr(Ka x 0.0125) = 4.7e-4 check to see if assumption is good 4.7e-4 / 0.0125) x 100 = 3.8% so good for SSA B) Based on previous theory adding strong base should lower the ionization %. i) 0.0125 | na | 0 | 0 c) -x | +x | +x e) 0.0125-x | 0.0025+x | x >> SSA allows both "-x" to be the same Ka = [(0.0025) x] / 0.0125 Solve for x = 9e-5 (9e-5 / 0.0125) x 100 = 0.72%

Given the titration curve, know the composition of various things (Press i) How protic?

Diprotic w/ equivalence points C and E A) H2A + H2O >< HA- + H3O+ (Ka1) [] [H2A] = Co, [HA-] = [H3O+] B) [H2A] = [AH-] (pKa1) C) HA- + H2O >< A2- + H3O+ (Ka2) HA- + H2O >< H2A + HO- (Kb) (=Kw/Ka1) D) HA- + NaOH --> A2- + Na+ + H2O >> Half second equivalence point >> [HA-] = [A2-] >> pH = pKa2 E) A2- + H2O >< HA- + HO- (Kb) (=Kw/Ka2) [A2-]= Co x (vi / vi+vi2eqiuv) >> [HA-]=[HO-]

Calculate how many moles of NaOH must be added to 0.5L of 0.2M HA to obtain a buffer solution with pH = 4. pKa = 4.20

HA + NaOH --> NaA + H2O nHA = 0.5L x 0.2M = 0.1mol mix 0.1 | x | na | na get 0.1-x | na | x pH = pKa + log([A-]/[HA]) 4.0 = 4.20 + log([A-]/[HA]) c=n/v [A-] = x / 0.5 [HA] = ((0.1-x)/0.5) >>0.5's cancel when in pH equation since dividng -0.2 = log (x / (0.1-x)) x=0.039mol -- Use HA leftover from neutralization and the A- formed from neutralizaiton

Write down mass and charge balance equations for 1.0×10−8mol/L HBr(aq). By considering the equations you wrote, which of the following expressions is correct? .A. [H3O+] = 1.0×10−8mol/L B.[H3O+] = 1.0×10−8mol/L + [OH−] C.[H3O+] = 1.1×10−8mol/L D.1.0×10−8mol/L = [H3O+] + [OH−] E.none of the expressions above is correct

HBr + H2O --> H3O+ + Br- MB: [HBr]o = [Br-] CB: [H3O+] = [Br-][OH-] >> Including self ionization

The pH of a dilute solution of HBr is 6.98 at 25oC (Assume HBr is 100% ionized) How much does the self-ionization of water contribute to the total amount of H3O+ in solution? A. (10e-7 / 10e-6.98)x100 = 95% B. (10e-7.02 / 10e-6.98)x100 = 91% C. (10e-6.98 / 10e-6.98)x100 = 100% D. (7.00-6.98)x100 = 2% -- At 25oC: - pH + pOH = 14.00 - Kw = 1e-14

HBr + H2O--> H3O+ + Br- H2O + H2O >< H3O+ + OH- Electrically neutral where [H3O+]=[OH-] + [Br-] HO- only from Kw H3O+ from both Ka and Kw pKw = pH + pOH pOH = 14-6.98 = 7.02 *B*

Titration curve of strong acid strong base? weak acid strong base? Differences? What is the half-point? Reaction? What abbout flip so adding strong acid to strong/weak base?

HCl + NaOH ---> H2O + NaCl Strongxstrong = lower initial pH; only water and salt so pH = 7 at equivalence point; mainly Na+ ad Cl- HA + NaOH --> H2O + NaA >> NaA is a weak base so only partially ionized, leaving some of it present at equivalence point to increase pH Weakxstrong = initial pH higher than in strong acid; equivalence point at pH>7 Equivalence point occurs at *same volume* Half point: Where [HA]=[NaA] = 1 ratio >>so using pH equation then *pH = pKa* >> can determine pKa Just flip 180o. SxS is pH still 7 at eq at wxS pH at eq is <7

What is/are the principal component(s) in solution at the equivalence point in the titration of HOCN(aq) with KOH(aq)? A.HOCN and OH− B.OCN−and K+ C.OCN−and OH− D.HOCN and OCN− E. H3O+and OH−

HOCN + HO- --> OCN- + H2O + Na+ At eq moles are equal Neutralization kills all HOCN and HO- Leftover are OCN- and K+

When do you think inductive and when size of anion?

Inductive if EN element is away that tugs the H Size if its an actual anion (e.g. OH-) that dissociates, consider stability of it

>> Describe the ionization of aweak acid and Le Chatelier's Principle How is % ionization of HA affected by: 1. Dilution? 2. Addition of H3O or A-? -- Does this differ with base dissociation?

Ka - ((nA-nH3O+) / nHA) x (1/V) %ionization increases as initial concentration decreases Adding more H2O reactants moves reaction right to compensate Adding H3O" (by HCl) or A-(by CH3COONa) triggers rxn left (minimizes ionization) due to stoich balance -- NH3 + H2O ><NH4+ + OH- NaOH --> Na+ + OH- A base eq will go left if a strong base is added (same stoich reasons) leading to a supressed ionization Adding a weak acid will also supressed ionization

*How many moles of NaOH must be added to 0.850 L of 0.179 mol/L HC3H5O2(aq) to obtain a solution with pH=5.1? Assume the temperature is 25oC. * (For HC3H5O2, pKa=4.89 at 25oC).

Ka = 10^-4.89 pKa = 4.89 pH = 5.1; [H+] = 10^-5.1 5.1 = 4.89 + ([NaOH]

If 50mL of 0.1M naOH and 20,: of 0.15 HA are mixed, then what is [HA] at equilibrium? Assume that HA is a weak monoproptic acid with pKa = 4.5 at 25oC

Ka = 10e-4.5 HA + NaOH --> NaA + H2O mol NaOH = 50mL x 0.1mol/L = 5mmol NaOH mol HA = 20mL x 0.15mol/L = 3mmol HA mixing: 3mmol | 5mmol | 0 | 0 >>HA limiting get: 0mmol | 2mmol | 3mmol | for total V of 70mL ice table for HA using "get values" i) 3mmol / 70mL | na | 0 | 2mmol / 70mL >>mmol/mL cancels to mol/L c) -x | +x | +x e) (3/70)-x | x | (2/70)+x Since conjugate acid base pair can do pKa + pKb = pKw or Kw = KaxKb Kb = 10e-14 / 10e-4.5 = 10e-9.5 >> can SSA now ignoring x when +/- it Kb = [x(2/70)] / (3/70) >>total volume (70) cancels x = 1.5 x 10e-9.5 = 4.74e-10M = [HA]

C6H5NH2(aq) + H2O(l) C6H5NH3+(aq) + HO-(aq)If 0.010 mol C6H5NH2 is dissolved in 0.50 L of water and the solution reaches equilibrium at room temperature, then which one of the following actions would cause net reaction to the right? A. Adding NaOH B.decreasing the temperature C.adding C6H5NH3Cl D.increasing the volume of the solution (by adding water) E.none of the above

NaOH inceases OH which pushes left COoling supports exo = left Salt moves left *D*

What is an equivalence point?

Point at which enough titrant has been added to react with all the acid (or base) nacid = nbase

What is a neutralization reaction? Concerted or multi-step? Eq large or small? How far will a rxn between weak vs strong species go?

Rxn between non conjugate species (acid and base) One way reaction (concerted) Kneut is very large Essentially to completion (at least one of two reactants is completely consumed (also with strong vs strong) -- Kw << Ka or Kb so Kneut >> 1

Effect of adding a strong acid or weak base into a acid dissociation reaction? Effect of adding a strong base or weak acid into a base dissociation

They will supress the ionizaiton as rxn goes left

Calculate the pH of a solution prepared by dissolving 0.015mol CH3COOH (Ka=1.8e-5) and 0.005mol NaOH in water ot make 1L of solution

Think neutralization first, consider moles CH3COOH + NaOH --> CH3COONa + H2O mixing: 15mmol | 5mmol | 0 | na get: 10mmol | - | 5 | na >> NaOH limiting since fewer moles >> use "get" to determine equilibrium in ice >> for 1L ice table for CH3COOH CH3COOH + H2O >< CH3COO- + H3O+ i) 0.01M | na | 0.005 | na ^^ 0.005 from neutralization complete dissociation >> net rxn right c) -x | +x | +x e) 0.01-x | 0.05+x | x = 0.01 | 0.05 | x Ka at e-5 is SSA Ka = 1.8e-5 = [(0.005)x] / 0.01 x = 2Ka = 3.6e-5 = [H3O+} pH = 4.44

Given the titration curve, know the composition of various things (Press i) Just based off picture, what is the unknown? Acid or base? What is [B]eq?

Unknown base A) B + H2O >< HB+ + HO- [] [B]eq = [B]o [] [HB]eq = [HO-] B) [] Half neutralization [] [B]eq = [HB+] [] pH = pKa C) [] nbase=nacid [] Ka = Kw/Kb = low [] [HB+]eq = Co x Vi/(Vi+VeP) = [Cl-] [] [H3O+] = [B] D) Excess HCl >> So no Base >> HCl determines pH HCl + H2O --> H3O+ +Cl- [Cl-] = [BH+] + [H3O+] Ka from HB+ + H2O >< B + H3O_ Solve for B known

Given the titration curve, know the composition of various things (Press i) Just based off picture, what is the unknown? Acid or base?

WEak Acid is unknown 1 eqiuvalence point = monoprotic A) [] HA + H2O >< A- + H3O+ [] Small Ka since weak acid so mainly just [HA] [] Equal [A-] and [H3O+] B) [] Half vol of equivalence point [] Only neutralzied half # moles [] [HA]eq = [A-]eq [] [HA]eq is not 1/2 initial since we added a base in so total volume is different!!! [] pH = pKa + log(A- / HA) [] pH = pKa >> buffer C) [] nbase = nacid [] since full neturalization since equal moles = [NaA] as main component [] [Na+] and [A-] []Where A- reacts with water, but low Kb so barely reacts with it ^^ Ka x Kb = Kw ^^ Kb = 10-9 so [] so [Na+] = [A-] [] [HA]eq = [HO-] D) Past eq so excess NaOH after neutralization (so no HA) [] [Na+] = [A-] + [OH-] >> Na comes from salt *and* NaOH = See next curve

Which of the following mixtures yields a buffer solution when dissolved in water to make1.0 L of solution? A.1.00 mol HBr and 0.60 mol NaBr B.1.00 mol CH3COOH and 0.60 mol HBr C.1.00 mol HBr and 1.00 mol NaOH D.1.00 mol CH3COOH and 0.60 mol NaBr E.1.00 mol CH3COOH and 0.60 mol NaOH

Weak acid and conjugate base in comparable and significant amounts buffer capacity max when ratio is 1:1 but works 1:10 or 10:1 A. HBr not weak B. Two acids C. NaOH is strong, HBr not weak D No not a pair, *E*. Yes, neut happens first, making weak base!!

What does pH=pKa mean? When does pH = pKa + 1? -1?

[conjugatebase] = [acid] = 1 = log(1) = 0 ratio is [B]:[A] = 10:1 ratio [B]:[A] = 1:10

Calculate the change in pH produced when a 5e-4mol HCl is added to a) 1L pure water at 25oC b) 1L of a buffer solution containing 0.01 mol CH3COOH and 0.005 CH3COONa pHbuffer = 4.44

a) pH of water at 25oC is 7 HCl + H2O --> Cl- + H3O+ I) 5.e-4 | na | F) 0 | na | 5e-4 | 5.e-4 [H3O+] = 5e-4M pH = 3.3 change in pH = -3.7 b) after adding acid, pH should go down (though by less) CH3COO- + HCl --> CH3COOH + NaCl mix: 5mmol | 0.5mmol | 10mmol get: 4.5mmol | -- | 10.5mmol CH3COOH + H2O >< CH3COO- + H3O+ i) 0.0105 | 0.0045| na c) -x | +x | +x e) 0.0105-x | 0.0045+x | x >> Can SSA Ka = 1.8e-5 = [(0.0045)x] / 0.0105 x=4.2e-5M [H3O+] pH = 4.43 change in pH = -0.06

What is the pH at the equivalence point when when 45.0 mL of 0.179 mol/L CH3COOH(aq) is titrated with 0.125 mol/L NaOH(aq)? Assume the temperature is 25oC. (For CH3COOH, pKa=4.74 at 25oC).

nCH3COOH = 45mL x 0.179mol/L = 8.055mmol vNaOH(mL) = 8.055mmol x 1 L / 0.125mol = 64.44mL NaOH + CH3COOH >< CH3COONa + H2O so Vtotal = 109.44mL mix. 8.055mmol | 8.055mmol --> 0 | na get. 0 | 0 -->8.055mmol | na We make a weak base (CH3COO- that dissociated from CH3COONa) CH3COO- + H2O >< CH3COOH + HO- i. 8.055mmol /109.44 | na | --> 0 | na c. -x | na | --> | +x | +x e. 8.055mmol /109.44 | x | x >> Cant SSA pKb = 14-4.74 = 9.26 Kb = 10^-9.26 10^-9.26 = x^2 / (8.055mmol /109.44) x = 6.36e-6 x = [HO-] pOH = 5.17 pH + pOH = pKw 14-5.17 = pH = *8.83*

What is the final pH of the solution when 15.0 mL of 0.127 mol/L NaOH(aq) and 35.0 mL of 0.183 mol/L HF(aq) are mixed? Assume the temperature is 25oC. (For HF, pKa=3.18 at 25oC).

nNaOH = 15mL x 0.127 mol/L = 1.905 mmol nHF = 35mL x 0.183mol/L = 6.405mmol vtotal = 50mL NaOH + HF >< NaF + H2O mix: 1.905mmol | 6.406mmol | --> | 0 | na get: 0 | 4.501 | --> | 1.905 | na -- F- + H2O >< HF + OH- i. (1.905/50) | na | --> | (4.501/50) | 0 c. -x | na | --> | +x | +x e. 0.0381-x | na | --> | 0.09002+x | x Kb = 14-3.18 = 10.82 = 10^-10.82 >> Can SSA e. 0.0381 | --> | 0.09002 | x Ka = x(0.09002) / 0.0381 x = 6.406e-12 = [OH-] pOH = 11.19 pH = *2.81* Constructing an ICE table allows the determination of the equilibrium concentration of H+. Remember that when the two solution are mixed the total volume of the solution will change!

What is the final pH at 25oC when 72.0 mL of 0.121 mol/L NaOH(aq) and 45.0 mL of 0.181 mol/L HA(aq) are mixed. Assume that HA is a weak monoprotic acid with pKa=5.39 at 25oC.

nNaOH = 72mL x 0.121mol/L = 8.712 mmol nHA = 45mL x 0.181mol/L = 8.145mmol vtot = 117mL HA + NaOH ><NaA + H2O mix. 8.145mmol | 8.712 mmol | --> | 0 | na get. 0 | 0.567 | --> | 8.145 | na We make a weak base (A- that dissociated from NaA) A- + H2O >< AH + HO- i. 8.145/117 | na | --> | 0 | 0 c. -x | na | --> | +x | +x e. (8.145/117)-x | x | nx pKb = 14-5.39 = 8.61 Kb = 10^-8.61 Can SSA e. (8.145/115) | x | x Kb = x^2 / (8.145/117) x = 1.31e-5 = [OH-] >> Negligible Consider excess NaOH NaOH + H2O >< Na + OH- So have 0.567 OH- moles [OH-] = 0.567/117 = 0.004846 pOH = 2.315 *pH = 11.69*

If 80.0 mL of 0.133 mol/L NaOH(aq) and 45.0 mL of 0.193 mol/L HA(aq) are mixed, then what is [HA] at equilibrium? Assume that HA is a weak monoprotic acid with pKa=5.29 at 25oC.

nNaOH = 80mL x 0.133mol/L = 10.64 mmol nHA = 45mL x 0.193mol/L = 8.685mmol vtot = 125mL HA + NaOH ><NaA + H2O mix. 8.685mmol | 10.64 mmol | --> | 0 | na get. 0 | 1.955 | --> | 8.685 | na We make a weak base (A- that dissociated from NaA) A- + H2O >< AH + HO- i. 8.685/125 | na | --> | 0 | 0 c. -x | na | --> | +x | +x e. (8.685/125)-x | x | x pKb = 14-5.29 = 8.71 Kb = 10^-8.71 Can SSA e. (8.685/125) | x | x Kb = x^2 / (8.685/125) x = *1.16e-5* = [AH] Since NaOH is in excess, initially only A- and OH- will be present, however A- can react with water to produce HA. It is valid to assume that, at equilibrium, x + [A-] ≈ [A-] and x + [OH-] ≈ [OH-] so long as x is much smaller than both [A-] and [OH-].

The pH of water 7.00 at 25oC and 6.13 at 100oC. These data indicate that, in water, A The acidity of water increases as temperatures increases B. [H3O+] increases as temperature increases C. Both the acidity and [H3O+] increase as temperature increases D. The acidity decreases as temperature increases E. [H3O+] decreases as temperature increases

pH = -log[H3O+] Pure water [OH-] = [H3O+] so pOH and pH are equal too This is irregardless of temperature Water is neutral at any tmperature To be acidic needs to pH < pOH, but acidity doesnt change pH = 1/2 x pKw >>in pure water So as pH decreases and temp increases, pKw decreases too so Kw goes up *B* What defines a neutral solution depends on the solution and the temperature (not always 7)

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pKa = 5.6 Ka = 10^-pKa Ka = 10^-5.6 HA + H2O >< A- + H3O+ i. 0.25 | na | 0 | 0 c. -x | na | +x | +x e. 0.25-x | x | x >> SSA since Ka low Ka = x^2 / 0.25 x = sqr(Ka x 0.25) x = 7.92e-4 check to see if assumption is good (7.92e-4 / 0.25) x 100 = *0.317%*