CHM 112 Exam 3

At 1 M, 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? - Extent of HNO2 ionization - The concentration of H+ - The pH

- Extent of HNO2 ionization decreases - The concentration of H+ decreases - The pH increases

Determines whether an aq solution will be acidic, basic, neutral

Acidic: cations of weak base, highly charged metal, anion strong acid (KCN) Basic: anions weak acid, cation strong base (FeCL3) Neutral: cation strong base, anoin strong acid (KNO3)

Concentration SO3^2- in eq with Ag2SO3 (s) and 1.80 x 10^-3 M Ag? Ksp will be provided

Ag2SO3 → 2Ag + SO3^2- Ksp = [Ag]^2 [SO3^2-]

Ka expression for CH3NH3 in water

CH3NH3+ (aq) + H2O (l) → H3O+ (aq) + CH3NH2 (aq) [H3O][CH3NH2] / [CH3NH3]

A buffer that contains 0.40 M base B, 0.25 M CA, BH+, has pH of 8.88. What is pH after 0.0020 mol HCl is added to 0.25 L of solution

Calculate mol BH+ and B Fill in ICE chart with mol BH+ and B in as initial Answer = C (produces additional x mol BH and consumes additional 0.0020 mol B)

Classify acid-base reactions Cu + 4Cl ← → CuCl4 Al(OH)3 + 3HNO3 ← → Al + 3H2O + 3NO3 N2 + 3H2 ← → 2 NH3 CN + H2O ← → HCN + OH

Cu + 4Cl ← → CuCl4 = Lewis Al(OH)3 + 3HNO3 ← → Al + 3H2O + 3NO3 = BL, A, Lewis N2 + 3H2 ← → 2 NH3 = not acid-base reaction CN + H2O ← → HCN + OH = BL, Lewis

Above what Fe conc will Fe(OH)2 precipitate from buffer pH = 9.52; ksp = 4.87 x10^-17

Fe(OH)2 ← → Fe + 2OH pOH = 14 - 9.52 = 4.48 [OH] = 10^-4.48 = 3.31 x 10^-5 Ksp = [Fe][OH]^2

pH of 3.48 mg/L Ba(OH)2 soln, complete dissociation?

Find concentration Ba(OH)2 (3.48 / molar mass) 1 mol Ba(OH)2 = 2 mol OH pOH = -log[OH] pH = 14 - pOH

pH of resulting soln if 34.0 mL of 0.340 M HCl is added to 44.0 mL of 0.340 M NaOH

HCl + NaOH → NaCl + H2O Excess mol NaOH = mol NaOH - mol HCl Total vol = vol HCl + vol NaOH [OH] = [NaOH] = excess NaOH/total vol pOH = -log[OH] pH = 14 - pOH

Common ion effect

HF (aq) + H2O (l) ← → H3O+ (aq) + F- (aq) Presence of common ion betw 2 solutes will cause eq to shift according to LC principle. NaF to HF eq shift left, away from excess F

Compound more soluble in water Strontium sulfate OR barium chromate Calcium carbonate OR copper (II) carbonate Barium iodate OR silver chromate

Incr. ksp = increased molar solubility if equal # ions Strontium sulfate > barium chromate Calcium carbonate > copper (II) carbonate Barium iodate > silver chromate

delta-pH when 3.00 mL of 0.100 M HCl is added to 100.0 mL buffer soln. 0.100 M in NH3 and 0.100 M in NH4Cl

Initial mol NH3 = Initial mol NH4Cl = 0.01 mol Initial pH = pka + log(NH3/NH4Cl) Addition of HCl: HCl + NH3 → NH4Cl Mol HCl added = 0.100(3/1000) = 0.0003 mol Mol NH3 = 0.01 - 0.0003 = 0.0097 mol Mol NH4Cl = 0.01 + 0.0003 = 0.0103 mol Final pH = pka + log(0.0097 / 0.0103) = 0.0261 decrease Addition of NaOH: NaOH + NH4Cl → NH3 + H2O + NaCl Mol NaOH added = 0.100(3/1000) = 0.003 mol Mol NH3 = 0.0103 mol Mol NH4Cl = 0.0097 mol Final pH = pka + log(0.0103/0.0092) Change in pH = pka + 0.0261 - pka = 0.0261 increase

Common ion expressions Iron (III) hydroxide Barium phosphate Tin(II) sulfide

Iron (III) hydroxide: ksp = [Fe3+][OH-]3 Fe(OH)3 (s) ← → Fe3+ (aq) - 3OH- (aq) Barium phosphate: ksp = [Ba2+]3[PO43-]2 Tin(II) sulfide: ksp = [Sn2+][HS-][OH-]

HF has a Ka of 6.8 x 10^-4. What are [H3O], [F], and [OH] in 0.75 M HF?

Ka = [H3O][F] / [HF] = (x)(x) / (0.75 - x), so x = 0.023 [H3O] = [F] = 0.023 M [OH] = 1 x 10^-14 / 0.023 = 4.4 x 10^-13 M

pH of 0.15 M methylammonium bromide, CH3NH3Br (kb CH3NH2 = 4.4 x 10^-4)

Ka = kw/kb = 2.27 x 10^-11 = (x)(x) / (0.15-x), so x = 1.85 x 10^-6 = [H3O] pH = -log[H3O] = 5.73

Higher pH? - Ka - pKa - Concentration - Acid strength - pOH

Lower Ka (increased k ionize, incr hydronium ions) Higher pKa (pKA = -log ka) Lower conc (Lower conc. = decr hydronium ions) Weak acid (vs. strong acid) Lower pOH (inversely related to pH)

Calculate pH during titration of 30 mL of 0.100 M KOH with 0.1 M HBr soln

Mol initial KOH 0 mL: pOH = -log(KOH) 15, etc. mL: mol added HBr = KOH that remains; pH = 14- pOH(molarity excess KOH) Repeat until 0 mol KOH remains. This is equivalence point, so pH = 7

pH of a vinegar w/ 5.0% acetic acid in water?

Molarity = (5.0 / 1 L)(1 mol / 60.5 g) = 0.833 Ka = 1.8 x 10^-5 = (x)(x) / (0.833 - x), so x = 3.87 = [H3O] pH = -log[H3O] = 2.41

Label acids, bases, conjugate pairs NH3 + HNO3 ← → NH4 + NO3 O + H2O ← → OH + OH NH4 + BRO3 ← → NH3 + HBrO3

NH3 (B) + HNO3 (A) ← → NH4 (CA) + NO3 (B) O (B) + H2O (A) ← → OH (CA) + OH (B) NH4 (A) + BRO3 (B) ← → NH3 (CB) + HBrO3 (CA)

Does any solid Ag2CrO4 form when 2.7 x 10^-5 g AgNO3 is dissolved in 15 mL of 4.0x10^-4M K2CrO4?

Qsp = [Ag]2[CrO4^2-] Qsp < Ksp, won't precipitate

NaOH with formic acid or acetic acid?

You want a buffer that is closer to desired pH. Calculate the pKa of each. NaOH partially neutralizes the acid and produces conjugate base.

Flask contains 0.572 g acid and few drops phenolphthalein indicator dissolved in water. Buret 0.180 M NaOH. vol base → endpoint = 25. Assuming acid is monoprotic.

molar mass = 0.572 g / (0.023 L * 0.180 M)

Calculate pH of buffer made by dissolving .8 mol NaOH in .5 L of 1.0 M H3PO4

pH = pKa + log(HPO4 / H2PO4) = -log(6.3x10^-8) + log((0.3/0.5) / (0.2/0.5)) = 7.38 Present initially is 0.5 mol H3PO4 0.5 mol H2PO4 + 0.3 mol NaOH 0.3 mol HPO4 + 0.2 mol H2PO4

pH when buffer soln is 0.110 M in weak acid (ka = 4.7 x 10^-5) and 0.530 in conj base

pH = pka + log [A]/[HA] = -log(4.7 x 10^-5) + log(0.530/0.110)

Indicator HA has ka = 1.3 x 10^-7

pKa = -log ka; at pH of 9 indicator > pKa so it ionizes

22*C excess M(OH)2 mixed with pure H2O; eq soln pH = 10.30

pOH = 14.10.30 = 3.70 [OH] = 10^-3.70 [M]=½ [OH] Ksp = [M][OH]^2 = 3.97x10^=12