At $90℃$, pure water has $\left\lbrack H^{+} \right\rbrack = 10^{- 6}$ M, if 100 mL of 0.2 M HCl is added to 200 mL of 0.1 M KOH at $90℃$ then pH of the resulting solution will be

Question

Dr. Abubacker · Accepted Answer

(d)
20 M eq. of HCl reacts completely with 20 Meq. Of KOH. Hence, the final solution is neutral. Thus, at $90℃$, pH = 6

Q. At $90℃$, pure water has $\left\lbrack H^{+} \right\rbrack = 10^{- 6}$ M, if 100 mL of 0.2 M HCl is added to 200 mL of 0.1 M KOH at $90℃$ then pH of the resulting solution will be

Explanation

Reviewed by Dr. Abubacker

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