Question

Calculate the pH of the solution that results from each of the following mixtures.

Part A

140.0 mL of 0.23 M  HF with 230.0 mL of 0.32 M  NaF

The Ka of HF is 6.6 x 10−4.

Express your answer using two decimal places.

Part B

185.0 mL of 0.12 M C2H5NH2 with 275.0 mL of 0.20 M C2H5NH3Cl

Express your answer using two decimal places.

Answer 1

A)

Concentration after mixing = mol of component / (total volume)

M(F-) after mixing = M(F-)*V(F-)/(total volume)

M(F-) after mixing = 0.32 M*230.0 mL/(230.0+140.0)mL

M(F-) after mixing = 0.1989 M

Concentration after mixing = mol of component / (total volume)

M(HF) after mixing = M(HF)*V(HF)/(total volume)

M(HF) after mixing = 0.23 M*140.0 mL/(140.0+230.0)mL

M(HF) after mixing = 8.703*10^-2 M

Ka = 6.6*10^-4

pKa = - log (Ka)

= - log(6.6*10^-4)

= 3.18

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.18+ log {0.1989/8.703*10^-2}

= 3.539

Answer: 3.54

B)

Concentration after mixing = mol of component / (total volume)

M(C2H5NH2) after mixing = M(C2H5NH2)*V(C2H5NH2)/(total volume)

M(C2H5NH2) after mixing = 0.12 M*185.0 mL/(185.0+275.0)mL

M(C2H5NH2) after mixing = 4.826*10^-2 M

Concentration after mixing = mol of component / (total volume)

M(C2H5NH3+) after mixing = M(C2H5NH3+)*V(C2H5NH3+)/(total volume)

M(C2H5NH3+) after mixing = 0.2 M*275.0 mL/(275.0+185.0)mL

M(C2H5NH3+) after mixing = 0.1196 M

Kb = 4.3*10^-4

pKb = - log (Kb)

= - log(4.3*10^-4)

= 3.367

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 3.367+ log {0.1196/4.826*10^-2}

= 3.761

use:

PH = 14 - pOH

= 14 - 3.7605

= 10.2395

Answer: 10.24