Calculate the pH of each of the solutions and the change in pH to 0.01 pH units

Question

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 3.84-M HCI to 530. mL of each of the following solutions. Change is defined as final minus initial, so if the pH drops upon mixing the change is negative. a) water pH before mixing = pH after mixing= pH change = b) 0.164 M C2H3O2^1- pH before mixing = pH after mixing= pH change = c) 0.164 M HC,H3O2 pH before mixing = pH after mixing= pH change = d) a buffer solution that s 0.164 M in each C,H302^1- and HC2H3O2 pH before mixing = pH after mixing= pH change =

Explanation / Answer

Given,

Conc. of HCl = 3.84 M

Volume of HCl added = 10 mL

=> Moles of HCl added = 3.84 x 0.01 = 0.0384 moles

Volume of Solutions = 530 mL

=> Final Total Volume = 530 + 10 = 540 mL = 0.54 L

=> Final [HCl] = 0.0384 / 0.54 = 0.0711 M

a) Water

pH before mixing = 7

We know that HCl is a Strong Acid and it dissociated completely in water to produce H+ and Cl-

HCl -----> H+ + Cl-

=> [H+] = [HCl] = 0.0711 M

pH = - log [H+] = - log 0.0711 = 1.15 = pH after Mixing

pH change = 1.15 - 7 = -5.85

b) 0.164 M CH3COO-

Kb for CH3COO- = 5.75 x 10^-10

CH3COO- + H2O -----> CH3COOH + OH-

0.164 - X.............................X..............X

Kb = 5.75 x 10^-10 = X^2 / 0.164 - X

=> X = 9.715 x 10^-6 M = [OH-]

pOH = - log [OH-]

=> pOH = 5.01

=> pH = 14 - 5.01 = 8.99 = pH before Mixing

CH3COO- + HCl ------> CH3COOH + Cl-

We have 0.0711 M HCl and 0.164 M CH3COO-

Therefore,

[CH3COOH] = 0.0711 M

[CH3COO-] = 0.164 - 0.0711 = 0.0929 M

It becomes a BUFFER

pH for a buffer = pKa + log (CH3COO- / CH3COOH)

pKa = 4.76

=> pH = 4.76 + log (0.0929 / 0.0711) = 4.88 = pH after mixing

pH change = 4.88 - 8.99 = -4.11

c) 0.164 M CH3COOH

Ka = 1.74 x 10^-5

CH3COOH ---------> CH3COO- + H+

0.164 - X....................X................X

Ka = 1.74 x 10^-5 = X^2 / 0.164 - X

=> X = 1.69 x 10^-3 M = [H+]

=> pH = 2.77 = pH before mixing

After mixing

[H+] from HCl = 0.0711 M

CH3COOH ---------> CH3COO- + H+

0.164 - X.....................X...........X + 0.0711

Ka = 1.74 x 10^-5 = (X) (X+0.0711) / (0.164 - X)

=> X = 4.01 x 10^-5 M

=> [H+] =0.0711 + X = 0.07114 M

=> pH = 1.15 = pH after mixing

=> pH chenge = 1.15 - 2.77 = -1.62

d) pH for a buffer = pKa + log (CH3COO- / CH3COOH)

=> pH = 4.76 + log (0.164 / 0.164)

=> pH = 4.76 = pH before mixing

pH after adding 0.0711 M HCl

pH = 4.76 + log (0.164 - 0.0711 / 0.164 + 0.0711) = 4.36 = pH after mixing

pH change = 4.36 - 4.76 = 0.4

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