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Calculate the $pH$ of a $4 \times 10^{-3} \ M \ Y(OH)_2$ solution. Assume $100\%$ dissociation for the first step and $50\%$ dissociation for the second step.

When $100 \ mL$ of $0.1 \ M$ $NaOH$ is added to $50 \ mL$ of $0.2 \ M$ $HCl$ at $25 \ ^oC$,the $pH$ of the resulting solution will be:

Given that the dissociation constant for $H_2O$ is $K_w = 1 \times 10^{-14} \ mol^2/L^2$. What is the $pH$ of a $0.001 \ M$ $KOH$ solution?

Which will have maximum $pH$?

The $pH$ of a $10^{-9} \ M$ solution of $HCl$ in water is

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