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Q. A sample of $ N{{a}_{2}}C{{O}_{3}}.{{H}_{2}}O $ weighing 0.62 g is added to 100 mL of $ 0.1\text{ }N\,{{(N{{H}_{4}})}_{2}}S{{O}_{4}} $ solution. What will be the resulting solution?

BHUBHU 2011

Solution:

Gram equivalent of $ {{(N{{H}_{4}})}_{2}}S{{O}_{4}} $
$=\frac{100}{1000}\times \frac{1}{10}\times 66=0.66 $
Gram equivalent of $ N{{a}_{2}}C{{O}_{3}}.{{H}_{2}}O $
$=\frac{0.62}{62}=0.01 $
Left $ {{(N{{H}_{4}})}_{2}}.S{{O}_{4}} $
is $ 0.66-0.01=0.65 $
$ {{(N{{H}_{4}})}_{2}}S{{O}_{4}} $
is salt of strong acid and weak base.