Question

A student dissolves 0.0100 mole of an unknown weak base in 100.00 mL water and titrates the

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 $1.499\times 10^{-6}$