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⁺](\(\frac{1}{221}\)) 
= 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