Question
A mixture of $$200\,mmol$$ of $$Ca{\left( {OH} \right)_2}$$ and $$4\,g$$ of sodium sulphate was dissolved in water and the volume was made up to $$100\,mL.$$ The mass of calcium sulphate formed and the concentration of $$O{H^ - }$$ in resulting solution respectively, are ( Molar mass of $$Ca{\left( {OH} \right)_2},N{a_2}S{O_4}$$ and $$CaS{O_4}$$ are $$74,143$$ and $$136\,g\,mo{l^{ - 1}},$$ respectively. )
A.
$$3.81\,g,0.56\,\,mol\,\,{L^{ - 1}}$$
B.
$$13.6\,g,0.56\,\,mol\,\,{L^{ - 1}}$$
C.
$$3.81\,g,0.14\,\,mol\,\,{L^{ - 1}}$$
D.
$$13.6\,g,0.14\,\,mol\,\,{L^{ - 1}}$$
Answer :
$$3.81\,g,0.56\,\,mol\,\,{L^{ - 1}}$$
Solution :
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\,\,\,\,\,Ca{{\left( {OH} \right)}_2}}\limits_{1\,\,mol} \,\,\,\,\,\, + \mathop {\,\,\,N{a_2}S{O_4}}\limits_{1\,\,mol} \,\,\,\,\, \to \mathop {\,\,CaS{O_4}}\limits_{1\,\,mol} + \mathop {2NaOH}\limits_{2\,\,mol} \cr
& {\text{Given,}}\,\,\,\,\,\,\,\,200\,\,mmol\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{4}{{142}}mol \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \approx 28\,\,mmol \cr} $$
$${\text{Thus,}}\,\,N{a_2}S{O_4}\,\,{\text{is limiting reagent}}{\text{.}}$$ $${\text{So,}}\,28\,\,mmol\,\,{\text{of}}\,\,CaS{O_4}\,\,{\text{will be produced}}{\text{.}}$$
$$28\,\,mmol\,\,{\text{of}}$$ $$CaS{O_4} \equiv 28 \times {10^{ - 3}} \times 136 = 3.81\,g$$
$$28\,\,mmol\,\,Ca{\left( {OH} \right)_2}\,\,{\text{will produce}}$$ $$ = 56\,\,mmol\,\,{\text{of}}\,\,NaOH.$$
$$\eqalign{
& \left[ {O{H^ - }} \right] = \frac{{56}}{{100}} \times 1000\frac{{mmol}}{L} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 560\,\,mmol/L = 0.56\,\,mol/L \cr} $$