Question

A student dissolves 0.0100 mole of an unknown weak base in 100.00 mL water and titrates the
solution with 0.100 MHNO3. After 40.00 mL of 0.100M HNO3 was added, the pH of the resulting solution was 8.00. Calculate Kb
value for the weak base.

Answer 1

Answer: The value of $K_b$ for weak base is $1.499\times 10^{-6}$

Explanation:

We are given:

Moles of unknown weak base = 0.0100 moles

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of nitric acid = 0.100 M

Volume of solution = 40.0 mL

Putting values in above equation, we get:

$0.100M=\frac{\text{Moles of nitric acid}\times 1000}{40mL}\\\\\text{Moles of nitric acid}=\frac{(0.100\times 40}{1000}=0.004mol$

The chemical reaction for unknown weak base (BOH) and nitric acid follows the equation:

                $BOH+HNO_3\rightarrow BNO_3+H_2O$

Initial:          0.0100       0.004        

Final:           0.006          -             0.004

Volume of solution = 100 + 40 = 140 mL = 0.140 L    (Conversion factor:  1 L = 1000 mL)

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_a+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[BNO_3]}{[BOH]})$

We are given:

$pK_b$ = negative logarithm of acid dissociation constant of formic acid = ?

$[BNO_3]=\frac{0.004}{0.140}$

$[BOH]=\frac{0.006}{0.140}$

pH = 8.00

pOH = 14 - pH = 14 - 8 = 6

Putting values in above equation, we get:

$6=pK_b+\log(\frac{0.006/0.140}{0.004/0.140})\\\\pK_b=5.824$

We know that:

$pK_b=-\log K_b$

$5.824=-\log K_b\\\\K_b=10^{-5.824}=1.499\times 10^{-6}$

Hence, the value of $K_b$ for weak base is $1.499\times 10^{-6}$