Question

What is the pH of a buffer solution upon mixing 15.0 mL of 0.40 M HCl and 20.0 mL of 0.50 M NH? Kb (NH3) = 1.8 x 10 E. 7.00 A. 9.43 B. 9.26 C. 9.08 D. 8.11

Answer 1

Answer: The pH of resulting solution is 9.08

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$       ........(1)

• For HCl:

Molarity of HCl = 0.40 M

Volume of solution = 15.0 mL

Putting values in equation 1, we get:

$0.40M=\frac{\text{Moles of HCl}\times 1000}{15.0mL}\\\\\text{Moles of HCl}=0.006mol$

• For ammonia:

Molarity of ammonia = 0.50 M

Volume of solution = 20.0 mL

Putting values in equation 1, we get:

$0.50M=\frac{\text{Moles of ammonia}\times 1000}{20.0mL}\\\\\text{Moles of ammonia}=0.01mol$

The chemical reaction for hydrochloric acid and ammonia follows the equation:

                  $HCl+NH_3\rightarrow NH_4Cl$

Initial:          0.006      0.01

Final:             -         0.004              0.006

Volume of solution = 15.0 + 20.0 = 35.0 mL = 0.035 L    (Conversion factor:  1 L = 1000 mL)

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

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

$pOH=pK_b+\log(\frac{[NH_4Cl]}{[NH_3]})$

We are given:

$pK_b$ = negative logarithm of base dissociation constant of ammonia = $-\log (1.8\times 10^{-5})=4.74$

$[NH_4Cl]=\frac{0.006}{0.035}$

$[NH_3]=\frac{0.004}{0.035}$

pOH = ?

Putting values in above equation, we get:

$pOH=4.74+\log(\frac{0.006/0.035}{0.004/0.035})\\\\pOH=4.92$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-4.92=9.08$

Hence, the pH of the solution is 9.08