Calculate the $pH$ of the following solutions:
$(a)$ $2 \,g$ of $TlOH$ dissolved in water to give $2 \,L$ of solution.
$(b)$ $0.3 \,g$ of $Ca(OH)_2$ dissolved in water to give $500 \,mL$ of solution.
$(c)$ $0.3 \,g$ of $NaOH$ dissolved in water to give $200 \,mL$ of solution.
$(d)$ $1 \,mL$ of $13.6 \,M \,HCl$ is diluted with water to give $1 \,L$ of solution.

(N/A) Molar mass of $TlOH = 204.4 + 16 + 1 = 221.4 \,g/mol$. Concentration $[TlOH] = \frac{2 \,g}{221.4 \,g/mol \times 2 \,L} \approx 0.004516 \,M$. Since $TlOH$ is a strong base,$[OH^-] = 0.004516 \,M$. $pOH = -\log(0.004516) \approx 2.345$. $pH = 14 - 2.345 = 11.655$.
$(b)$ Molar mass of $Ca(OH)_2 = 40 + 2(16 + 1) = 74 \,g/mol$. Moles of $Ca(OH)_2 = \frac{0.3 \,g}{74 \,g/mol} \approx 0.004054 \,mol$. Concentration $[Ca(OH)_2] = \frac{0.004054 \,mol}{0.5 \,L} = 0.008108 \,M$. Since $Ca(OH)_2$ provides $2 \,OH^-$ ions,$[OH^-] = 2 \times 0.008108 = 0.016216 \,M$. $pOH = -\log(0.016216) \approx 1.79$. $pH = 14 - 1.79 = 12.21$.
$(c)$ Molar mass of $NaOH = 23 + 16 + 1 = 40 \,g/mol$. Moles of $NaOH = \frac{0.3 \,g}{40 \,g/mol} = 0.0075 \,mol$. Concentration $[NaOH] = \frac{0.0075 \,mol}{0.2 \,L} = 0.0375 \,M$. $[OH^-] = 0.0375 \,M$. $pOH = -\log(0.0375) \approx 1.426$. $pH = 14 - 1.426 = 12.574$.
$(d)$ Using dilution formula $M_1V_1 = M_2V_2$: $13.6 \,M \times 1 \,mL = M_2 \times 1000 \,mL$. $M_2 = 0.0136 \,M = [H^+]$. $pH = -\log(0.0136) \approx 1.866$.