question_answer 1)

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 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 litre of \[HCl\]solution.  

Answer:

(a) Molarity, \[M=\frac{{{w}_{B}}\times 1000}{{{m}_{B}}\times V}\] where\[{{w}_{B}}\] = mass of solute \[{{m}_{B}}\] = molar mass of solute V = volume of solution \[M=\frac{2\times 1000}{204\times 2000}=4.9\times {{10}^{-3}}M\] \[TlOH(aq)\to T{{l}^{+}}+O{{H}^{-}}(aq)\] \[[O{{H}^{-}}]=4.9\times {{10}^{-3}}M\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log [4.9\times {{10}^{-3}}]\] \[=2.309\] \[pH=14-2.309=11.691\] (b) Molarity of \[Ca{{(OH)}_{2}},M=\frac{{{w}_{B}}\times 1000}{{{m}_{B}}\times V}\] \[=\frac{0.3\times 1000}{74\times 500}\] \[=8.1\times {{10}^{-3}}M\] \[Ca{{(OH)}_{2}}\to C{{a}^{2+}}(aq)+2O{{H}^{-}}(aq)\] \[[O{{H}^{-}}]=2\times 8.1\times {{10}^{-3}}=0.0162M\] \[pOH=-{{\log }_{10}}[O{{H}^{-}}]\] \[=-\log [0.0162]=1.79\] \[pH=14-1.79=12.21\] (c) Molarity of \[NaOH,M=\frac{{{w}_{B}}\times 1000}{{{m}_{B}}\times V}\] \[=\frac{0.3\times 1000}{40\times 200}=0.0375\] \[NaOH(aq)\to N{{a}^{+}}(aq)+O{{H}^{-}}(aq)\] \[[O{{H}^{-}}]=0.0375M\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log [0.0375]\] \[=1.4259\] \[pH=14-1.4259=12.57\] (d) \[{{M}_{1}}{{V}_{1}}\] (before dilution) = \[{{M}_{2}}{{V}_{2}}\] (after dilution) \[13.6\times 1={{M}_{2}}\times 1000\] \[{{M}_{2}}=0.0136M\] \[HCl(eq)\to {{H}^{+}}(aq)+C{{l}^{-}}(aq)\] \[[{{H}^{+}}]=0.0136M\] \[pH=-\log [0.0136]\] =1.866