JEE MAIN - Chemistry (2021 - 31st August Evening Shift - No. 17)
The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is x $$\times$$ 10$$-$$4. The value of x is ____________. (Nearest integer) [log 2.5 = 0.3979]
Answer
6021
Explanation
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$$[HCl] = {{20} \over {80}} = {1 \over 4}M = 2.5 \times {10^{ - 1}}M$$
pH = $$-$$log 2.15 $$\times$$ 10-1 = 1 $$-$$ 0.3979 = 0.6021
pH = 6021 $$\times$$ 10-4
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