Question

: Calculate the pH of a solution that results from 125 mL of 0.10M NH3 with 250 mL of 0.10 M NH4Cl; Kb = 1.76 x 10-5

Calculate the pH when 0.010 mols of NaOH are added to 250.0 mL buffer containing 0.250 M acetic acid and 0.250 M sodium acetate. (pka = 4.74)

A 50.0 mL sample of 0.200M NaOH is titrated with 0.200M nitric acid. Calculate the pH after 25.0 mL HNO3 has been added.

Answer 1

1)

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

M(NH3) after mixing = M(NH3)*V(NH3)/(total volume)

M(NH3) after mixing = 0.1 M*125.0 mL/(125.0+250.0)mL

M(NH3) after mixing = 3.333*10^-2 M

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

M(NH4+) after mixing = M(NH4+)*V(NH4+)/(total volume)

M(NH4+) after mixing = 0.1 M*250.0 mL/(250.0+125.0)mL

M(NH4+) after mixing = 6.667*10^-2 M

Kb = 1.76*10^-5

pKb = - log (Kb)

= - log(1.76*10^-5)

= 4.754

use:

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

= 4.754+ log {6.667*10^-2/3.333*10^-2}

= 5.056

use:

PH = 14 - pOH

= 14 - 5.0555

= 8.9445

Answer: 8.94

2)

mol of NaOH added = 0.01 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 0.25 M *0.25 L

mol of CH3COO- = 0.0625 mol

mol of CH3COOH = 0.25 M *0.25 L

mol of CH3COOH = 0.0625 mol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (0.0625 + 0.01) mol

mol of CH3COO- = 0.0725 mol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (0.0625 - 0.01) mol

mol of CH3COOH = 0.0525 mol

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

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

= 4.74+ log {7.25*10^-2/5.25*10^-2}

= 4.88

Answer: 4.88

3)

Given:

M(HNO3) = 0.2 M

V(HNO3) = 25 mL

M(NaOH) = 0.2 M

V(NaOH) = 50 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.2 M * 25 mL = 5 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.2 M * 50 mL = 10 mmol

We have:

mol(HNO3) = 5 mmol

mol(NaOH) = 10 mmol

5 mmol of both will react

remaining mol of NaOH = 5 mmol

Total volume = 75 mL

[OH-]= mol of base remaining / volume

[OH-] = 5 mmol/75 mL

= 6.667*10^-2 M

use:

pOH = -log [OH-]

= -log (6.667*10^-2)

= 1.1761

use:

PH = 14 - pOH

= 14 - 1.1761

= 12.8239

Answer: 12.82