question_answer 94)

Calculate the pH of the resultant mixtures: (a)\[10\,mL\,of\,0.2MCa{{(OH)}_{2}}+25mL\,of0.1M\,HCl\](b) \[10mL\,of\,0.1\,M\,{{H}_{2}}S{{O}_{4}}+10mL\,of\,0.01M\,Ca{{(OH)}_{2}}\] (c)\[10mL\,of\,0.1\,M\,{{H}_{2}}S{{O}_{4}}+10mL\,of\,0.1M\,KOH\]  

Answer:

(a) Number of moles of \[Ca{{(OH)}_{2}}=\frac{MV}{1000}=\frac{0.2\times 10}{1000}=0.002\] Number of moles of \[HCl=\frac{MV}{1000}=\frac{0.1\times 25}{1000}=0.0025\] \[{{n}_{O{{H}^{-}}}}=2\times 0.002=0.004\,\,\,\,;\,\,\,{{n}_{{{H}^{+}}}}=0.0025\] \[{{H}^{+}}+O{{H}^{-}}\to {{H}_{2}}O\] Number of remaining moles of OH- ions after neutralisation = 0.004 - 0.0025 = 0.0015 Molarity of \[O{{H}^{-}}\] ion = \[\frac{{{n}_{O{{H}^{-}}}}\times 1000}{V}=\frac{0.0015\times 1000}{35}\] \[=0.0428\] \[pOH=-\log \,\,[O{{H}^{-}}]\] \[=-\log [0.0428]\] \[=1.3679\] \[pH=14-1.3679=12.6321\] (b) Number of moles of \[{{H}^{+}}\] ions = 2 x number of moles of \[{{H}_{2}}S{{O}_{4}}\] \[=2\times \frac{MV}{1000}=\frac{2\times 0.01\times 10}{1000}=2\times {{10}^{-4}}\] Number of moles of \[O{{H}^{-}}\] ions = 2 x number of moles of \[Ca{{(OH)}_{2}}\] \[=2\times \frac{MV}{1000}=\frac{2\times 0.01\times 10}{1000}=2\times {{10}^{-4}}\] Since, number of moles of \[{{H}^{+}}\] and \[O{{H}^{-}}\] ions are same hence they will neutralise each other and the pH of solution will be 7. (c) Number of moles of \[{{H}^{+}}\]ions = 2 x number of moles of \[{{H}_{2}}S{{O}_{4}}\] \[=2\times \frac{MV}{1000}\] \[=2\times \frac{0.1\times 10}{1000}=2\times {{10}^{-3}}\] Number of moles of \[O{{H}^{-}}\] ions = number of moles of KOH \[=\frac{MV}{1000}=\frac{0.1\times 10}{1000}={{10}^{-3}}\] \[{{H}^{+}}+O{{H}^{-}}\xrightarrow{{}}{{H}_{2}}O\] Remaining moles of \[{{H}^{+}}\]ions after neutralization \[=2\times {{10}^{-3}}-1\times {{10}^{-3}}-1\times {{10}^{-3}}=1\times {{10}^{-3}}\] Molarity of \[{{H}^{+}}\] ions\[=\frac{n}{V}\times 1000\] \[=\frac{{{10}^{-3}}\times 1000}{20}=0.05\] \[pH=-{{\log }_{10}}\,[{{H}^{+}}]=-{{\log }_{10}}[0.05]=1.301\]