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Exercises · 6.49

Q.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.

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For strong bases find [OH−][\text{OH}^-], then pH=14−pOH\text{pH} = 14 - \text{pOH}; for the strong acid find [H+][\text{H}^+] directly. Results: (a) 11.65, (b) 12.21, (c) 12.57, (d) 1.87.

a) 2 g TlOH in 2 L

M(TlOH)=204.4+17.0=221.4 g mol−1M(\text{TlOH}) = 204.4 + 17.0 = 221.4\ \text{g mol}^{-1}

[TlOH]=2/221.42=4.52×10−3 M=[OH−][\text{TlOH}] = \frac{2/221.4}{2} = 4.52\times10^{-3}\ \text{M} = [\text{OH}^-]

pOH=−log⁡(4.52×10−3)=2.35 ⇒ pH=14−2.35=11.65\text{pOH} = -\log(4.52\times10^{-3}) = 2.35 \ \Rightarrow\ \text{pH} = 14 - 2.35 = 11.65

b) 0.3 g Ca(OH)₂ in 500 mL

M(Ca(OH)2)=40.1+2(17.0)=74.1 g mol−1M(\text{Ca(OH)}_2) = 40.1 + 2(17.0) = 74.1\ \text{g mol}^{-1}

[Ca(OH)2]=0.3/74.10.5=8.10×10−3 M[\text{Ca(OH)}_2] = \frac{0.3/74.1}{0.5} = 8.10\times10^{-3}\ \text{M}

Each formula unit gives 2 OH−\text{OH}^-: [OH−]=2(8.10×10−3)=1.62×10−2 M[\text{OH}^-] = 2(8.10\times10^{-3}) = 1.62\times10^{-2}\ \text{M}

pOH=−log⁡(1.62×10−2)=1.79 ⇒ pH=12.21\text{pOH} = -\log(1.62\times10^{-2}) = 1.79 \ \Rightarrow\ \text{pH} = 12.21

c) 0.3 g NaOH in 200 mL

M(NaOH)=40.0 g mol−1M(\text{NaOH}) = 40.0\ \text{g mol}^{-1} …

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