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  4. The pH of solution formed by mixing 40 mL of 0.10 M HCl with 10 mL of 0.45 M of NaOH is

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Q. The pH of solution formed by mixing 40 mL of 0.10 M HCl with 10 mL of 0.45 M of NaOH is

Equilibrium

Solution:

Millimoles of $HCl=40\times0.1=4;$
Millimoles of $NaOH = 10\times 0.45 = 4.5$
Millimoles of $NaOH$ left $= 0.5$
$\therefore $ Volume of solution $=50$ mL
$\therefore $ Moles of $NaOH$ per litre
$=\frac{0.5 \times 10^{-3} \times 1000}{50}=1\times10^{-2}\,M$
$\therefore $ pOH of solution $=-log (1 \times 10^{-2})=2$
$\therefore pH=14-2=12$