Question

A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 13.0 mL of KOH. Express your answer numerically.

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 15.0 mL of HNO3.

Express your answer numerically.

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 33.0 mL of NaOH.

Express your answer numerically

Answer 1

Solution :-

A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 13.0 mL of KOH. Express your answer numerically.

Solution :- HBr and KOH are strong acid and strong base

Lets first calculate the moles of each using the molarity and volume

Moles = molarity * volume in liter

Moles of HBr = 0.15 mol per L * 0.050 L = 0.0075

Moles of KOH = 0.250 mol per L * 0.013 L = 0.00325

Moles of KOH are less therefore lets calculate the moles of HBr remain after the reaction

Moles of HBr remain = 0.0075 – 0.00325 = 0.00425 mol HBr

Now lets calculate the new molarity of HBr at total volume (50.0 + 13.0 = 63.0 ml = 0.063 L)

[HBr] = moles / volume in liter

           = 0.00425 mol / 0.063 L

          = 0.06746 M

Now lets calculate the pH

pH= - log [H+]

pH= - log [0.06746]

pH= 1.17

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 15.0 mL of HNO3.

Solution :- lets first calculate the moles of the each

Moles of NH3 = 0.200 mol per L * 0.075 L = 0.015 mol

Moles of HNO3 = 0.500 mol per L * 0.0150 L = 0.0075 mol

Moles of HNO3 are less so after the reaction conjugate acid is formed equal to the moles of the HNO3 and moles of NH3 decreased by the same moles of HNO3

So after the reaction moles of NH3 = 0.015 – 0.0075 = 0.0075 mol NH3

Moles of NH4^+ = 0.0075 mol

Using the Henderson equation we can calculate the pOH

pOH= pkb + log [acid ]/[base]

pOH = 4.74 + log [0.0075/0.0075]

pOH = 4.74

pH+ pOH = 14

therefore

pH= 14 – 4.74

pH= 9.26

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 33.0 mL of NaOH.

Solution :-

Lets first calculate the moles of each

Moles of CH3COOH= 0.35 mol per L * 0.052 L = 0.0182 mol

Moles of NaOH = 0.40 mol per L * 0.033 L = 0.0132 mol

Moles of acetic acid after the reaction = 0.0182 – 0.0132 = 0.005

Moles of acetate after reaction = 0.0132 mol

Now lets calculate the concentration of each at total volume

52 ml + 33 ml = 85 ml = 0.085 L

[CH3COOH] = 0.005 mol / 0.085 L = 0.0588 M

[CH3COO-] = 0.0132 mol / 0.085 L = 0.155 M

Now lets calculate the pH using the Henderson equation

pH= pka + log ([base]/[acid])

pH= 4.74 + log [0.0588/0.155]

pH= 4.32