Question

$8gNaOH$ and $4.9g{H}_{2}S{O}_4$ are present in one litre of the solution. What is its $pH$?

• A. $1$
• B. $13$
• C. $12$
• D. $2$

Answer 1

Answer: (B)

$13$

Molecular weight of NaOH = 40

Weight of NaOH = 8g

Moles of NaOH = Weight/Molecular weight = 8/40 = 0.2 moles

Molecular weight of $H_{2}S{O}_4$ = 98

Weight of $H_{2}S{O}_4$ = 4.9 g

Moles of $H_{2}S{O}_4$ = $\frac{Weight}{Molecularweight}$ = =4.9/98 = 0.05 moles

1 mole of NaOH will give = 1 mole of $O{H}^{⁻}$ions

0.2 moles of NaOH will give = 0.2 moles of $O{H}^{⁻}$ ions

1 mole of $H_{2}S{O}_4$ will give = 2 moles of $H^{+}$ ions

0.05 moles of $H_{2}S{O}_4$ will give = 0.05 × 2 = 0.1 moles of $H^{+}$ ions

Thus 0.1 moles of $H^{+}$ will be neutralised by 0.1 moles of OH⁻ ions

And 0.1 moles of $O{H}^{-}$ ions will be left

Concentration of $O{H}^{-}$ ions = 0.1/1 = 0.1 M

or, [ $O{H}^{-}$ ] = 0.1 M

pOH = -log[$O{H}^{-}$ ] = -log(0.1) = -log(${10}^{-1}$) = 1

We know that,

pH + pOH = 14

Thus, pH + 1 = 14

or, pH = 14 - 1 = 13

Therefore pH of the solution is 13.

Hence , option B is correct .