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What is the [H^+] concentration of a 0.100 M CH_3CO_2H solution with K_a = 1.8 t

ID: 525355 • Letter: W

Question

What is the [H^+] concentration of a 0.100 M CH_3CO_2H solution with K_a = 1.8 times 10^-5 M? The equation for the dissociation of acetic acid is: CH_3CO_2H(aq) H^+(aq) + CH_3CO_2^-(aq) A) 4.2 times 10^-2 M B) 1.3 times 10^-2 M C) 1.3 times 10^-3 M D) 4.2 times 10^-3 M Calculate the pH of a solution that is 0.111 M in sodium formate (NaHCO_2) and 0.253 M in formic acid (HCO_2H). The K_a of formic acid is 1.77 times 10^-4 M. A) 10.61 B) 4.103 C) 3.387 D) 5.296 E) 14.36 A 100.0 mL sample of 0.18 M HCIO_4 is titrated with 0.27 M KOH. Determine the pH of the solution after the addition of 50.0 mL of KOH. A) 2.35 B) 12.48 C)3.22 D) 1.52 E) 0.68 What is the molar solubility of silver chloride ( AgCl) in water? K_sp for AgCl is 1.8 times 10^-10 M^2 at 25 degree C. A) 9.74 B) 1.9 times 10^-5 C) 9.0 times 10^-11 D) 3.6 times 10^-10 E)1.3 times 10^-5 The acid-dissociation constant of hydrocyanic acid (HCN) at 25.0 degree C is 4.9 timers 10^-10 M. What is the pH

Explanation / Answer

17) equilibrium concentration of weakness acid = (Ka*C)

[H^+] = (1.8x10^-5 x 0.1) => 1.34 x 10^-3 ( answer option C)

18) pH = pKa + log [ Na formate / formic acid]

pKa = - log (1.77 x 10^-4) => 3.75

pH = 3.75 + log [ 0.111 / 0.253] => 3.387 ( answer option C)

19) 100 mL x ( 0.18 mol /1000 mL) = 0.018 mol Of HClO4

50 mL x ( 0.27 mol / 1000 mL) = 0.0135 mol KOH

After mixing total volume = 150 mL

HClO4 + KOH = KClO4 + H2O

Remaining HClO4 = 0.018 - 0.0135 => 0.0045 mol in 150 mL

[H^+] = 0.0045 mol / 0.150 L => 0.03

pH => - log (0.03) = 1.52 ( option D)

20) AgCl (s) <==> Ag^+(aq) + Cl^-(aq)

Ksp =[Ag^+][Cl^-]

Ksp = x*x

x^2 = 1.8*10^-10

x = 1.3 *10^-5 (option E)

21) pH of aq salt composed by weak acid and strong base

pH = 7 + 1/2 pKa + 1/2 log C

pKa = - log (4.9 x 10^-10) => 9.30

C = 0.1M

pH = 7 + 1/2 *(9.30) + 1/2 log (0.100)

pH = 11.15 ( option B)

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