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\[=\frac{n}{V}\times 1000\]
\[=\frac{{{10}^{-3}}\times 1000}{20}=0.05\]
\[pH=-{{\log }_{10}}\,[{{H}^{+}}]=-{{\log }_{10}}[0.05]=1.301\]
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