Two solutions A and B, each of 100 L was made by dissolution of 4 g of NaOH and 9.8 g of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Equilibrium

Two solutions A and B, each of 100 L was made by dissolution of 4 g of NaOH and 9.8 g of

H_{2} SO_{4}

in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is

Answer: 10.6

Both stock solutions are

1 0^{- 3} M

\frac{4}{40 \times 100}

for NaOH and

\frac{9.8}{98 \times 100}

for

H_{2} S O_{4}

. Mixing

40 L

of base with

10 L

of acid supplies

40 \times 1 0^{- 3}

mol of

O H^{-}

against

10 \times 1 0^{- 3} \times 2 = 20 \times 1 0^{- 3}

mol of

H^{+}

, leaving

20 \times 1 0^{- 3}

mol of

O H^{-}

50 L

. Then

[O H^{-}] = 4 \times 1 0^{- 4} M

, so

pOH = - lo g (4 \times 1 0^{- 4}) = 3.4

and

pH = 14 - 3.4 = 10.6

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Numerical answer