Question

Calculate the pH of the following solutions:
a) 2 g of TlOH dissolved in water to give 2 litre 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) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution

Answer 1

(a) For 2g of TlOH dissolved in water to give 2 L of solution : [TIOH(aq)] =$\frac{2}{2}$ g / L = $\frac{2}{2}$ × $\frac{1}{221}$ M = $\frac{1}{221}$ M
TIO(aq) $\rightarrow$ TI + (aq) + OH-(aq)
= [OH-]
= [TIOH(aq)]
= $\frac{1}{221}$ M
KM = [H+][OH-]
10-14 = H+
= 221 × 10-14
= [H+] ⇒ pH
= -log [H+]
=-log (221 × 10-14)
= -log (2.21 ×10-12)
= 11.65

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(b) For 2g of TlOH dissolved in water to give 2 L of solution: Ca(OH2) $\rightarrow$ Ca2+ + 2OH-
= [Ca(OH)2]
= $\frac{0.3\times 1000}{500}$
= 0.6M
= [OH-aq] = 2 × [Ca(OH)2aq]
= 2 × 0.6 = 1.2M
[H+] = $\frac{KM}{[OH-aq]}$
= $\frac{10^{-14}}{1.2}$M
= 0.833 × 10-14
pH = -log(0.833 × 10-14)
= -log(8.33 × 10-13)
= (-0.902 + 13)
= 12.098

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(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution: NaOH $\rightarrow$ Na+ (aq) + OH-(aq)
= [NaOH]
= $\frac{0.3\times 1000}{200}$
= 1.5M
= [OH-aq]
= 1.5M
Then, [H+]
= $\frac{10-14}{1.5}$
= 6.66 × 10-13
= pH
= -log(6.66 × 10-13)
= 12.18

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(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution : 13.6 × 1 mL = M2 × 1000mL (Before dilution) (After dilution)
= 13.6 × 10-3
= M2 × 1L = M2
= 1.36 × 10-2
= [H+]
= 1.36 × 10-2
= pH = - log (1.36 × 10-2)
= (0.1335 + 2)
= 1.866 ~ 1.87