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

Step-by-step explanation:

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

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 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}* 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}* 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(([salt])/([base]))`

`pOH=pK_b+\log(([NH_4Cl])/([NH_3]))`

We are given:

`pK_b` = negative logarithm of base dissociation constant of ammonia =
`-\log (1.8* 10^(-5))=4.74`

`[NH_4Cl]=(0.006)/(0.035)`

`[NH_3]=(0.004)/(0.035)`

pOH = ?

Putting values in above equation, we get:

`pOH=4.74+\log((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