Question

100 ml of 0.1 mol/dm3 NaOH solution is added to 50ml of 0.1 mol/dm3 H2SO4 solution and filled with distilled water upto 250 ml level. What would be the final OH concentration in the solution?

Answer 1

Answer on Question #54138 – Chemistry – General chemistry

Question:

100 ml of 0.1 mol/dm³ NaOH solution is added to 50 ml of 0.1 mol/dm³ H₂SO₄ solution and filled with distilled water up to 250 ml level. What would be the final OH concentration in the solution?

Answer:

The neutralization reaction can be shown:

$$2 \mathrm{NaOH} + \mathrm{H_2SO_4} \rightarrow \mathrm{Na_2SO_4} + 2 \mathrm{H_2O}$$

It is clear that two moles of the base reacts with one mole of the acid; therefore the given amount of the reagents is equimolar:

$$\mathrm{C_1V_1} = 2 \mathrm{C_2V_2},$$

where $C_1$ and $V_1$ – the concentration and the volume of NaOH, and $C_2$ and $V_2$ – the concentration and the volume of $\mathrm{H_2SO_4}$.

$$100 \mathrm{ml} \times 0.1 \mathrm{mol/L} = 2 \times (50 \mathrm{ml} \times 0.1 \mathrm{mol/L})$$ $$10 \mathrm{mmol} = 2 \times 5 \mathrm{mmol}$$

Thus, the solution is neutral with pH of 7.

$$\mathrm{pOH} = 14 - \mathrm{pH} = 14 - 7 = 7$$

The final concentration of hydroxide anions equals:

$$[\mathrm{OH^-}] = 10^{-7} \mathrm{mol/L}$$

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