Question

Equal volume of 0.2M NH4OH and 0.1M H2SO4 are mixed.Calculate pH of the final solution?

Answer 1

pH = 9.30

Explanation:

According to Henderson - Hasselbalch equation:

pOH = pKb + log ([salt]/ [base])

Therefore, pOH = 5 + log (0.1 / 0.2)

= 5 + (-0.30) = 4.70

pH = 14 – pOH = 14 – 4.70

pH = 9.30

Answer 2

The pH of the final solution will be 7

Explanation:

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

$\text{Moles}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$     .....(1)

Molarity of $NH_4OH$ solution = 0.2 M

let Volume of $NH_4OH$ solution = 100 mL = 0.1 L

$\text{Moles of} NH_4OH={0.2}\times{0.1}=0.02moles$

Molarity of $H_2SO4$ solution = 0.1 M

let Volume of $H_2SO4$ solution = 100 mL = 0.1 L

$\text{Moles of} H_2SO4={0.1}\times{0.1}=0.01moles$

$2NH_4OH+H_2SO_4\rightarrow (NH_4)_2SO_4+H_2O$  

According to stoichiometry :

2 moles of $NH_4OH$ require 1 mole of $H_2SO_4$

Thus 0.02 moles of $NH_4OH$ will require=$\frac{1}{2}\times 0.02=0.01moles$  of $H_2SO_4$

As all the given amount of $NH_4OH$ will be completely used by the given amount of $H_2SO_4$ , the solution will be neutral and pH will be 7.

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