Question

A sample of hard water is found to contain 40mg of $Ca^{2+}$ ions per litre. The amount of washing soda $(Na_2CO_3)$ required to soften 5.0 L of the sample would be

• A. 1.06 g
• B. 5.3 g
• C. 53 mg
• D. 530 mg

Answer 1

Answer: (D)

The equation is
$Ca^{2+}+Na_{2}CO_{3} \to CaCO_{3}+2Na^{+}$
Total $Ca^{2+}$ in $5\,L$ of hard water $=5 \times 40=200\,mg$
Now from equation, $40\,g\,Ca^{2+}$ require washing sogd $=106\,g$
$\therefore 200\,mg$ of $Ca^{2+}$ will require washing soda
$=\frac{106}{40}\times200\times10^{-3}g$
$=0.530\,g$ or $530\,mg$