What volume of $0.05 M C a (O H)_{4}$ solution is needed for complete conversation of $10 m L$ of $0.1 M H_{2} P O_{4}$ into $C a (H_{2} P O_{4})_{2}$

Open in App

Solution

n-factor of $C a (O {H)}_{4}$ = 4 (because when dissociated, it gives out 4 OH⁻ ions)
$N o r m a l i t y = M o l a r i t y \times n - f a c t o r = 0.05 \times 2 = 0.1 = N_{1}$
Volume of $C a {O H}_{4}$ required $= V_{1}$
Normality of $H_{2} P O_{4}$ $= 0.1 = N_{2}$
Volume of $H_{2} P O_{4}$ $= 10 m l = V_{2}$
Using the formula $N_{1} V_{1} = N_{2} V_{2}$
$(0.1) \times (V_{1}) = (0.1) \times (10)$
$V_{2} = \frac{1}{0.1}$
$V_{2} = 10 m l$
Thus, 10ml of 0.05M $C a (O {H)}_{4}$ is required to neutralise 10ml of 0.1N $H_{2} P O_{4}$ .

Suggest Corrections

Similar questions

0.05 M C a (O H)_{2}

(i n m L)

10 m L

0.1 M H_{3} P O_{4}

C a (H_{2} P O_{4})_{2} ?

0.05 M C a (O H)_{2}

(i n m L)

10 m L

0.1 M H_{3} P O_{4}

C a (H_{2} P O_{4})_{2} ?

K M n O_{4}

100 m l

N a_{2} C O_{3}

N a H C O_{3}

H C l

0.05 M K_{2} C r_{2} O_{7}

200

0.6 M

F e C_{2} O_{4}

Join BYJU'S Learning Program