What will be the $p H$ of a solution formed by mixing $40 m L$ of $0.10 M H C l$ with $10 m L$ of $0.45 M N a O H$ ?

12

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $12$
Millimoles of $H C l = 40 \times 0.10 M = 4 m m o l$
Millimoles of $N a O H = 10 \times 0.5 = 4.5 m m o l$
Since, $N a O H$ is in excess,
Millimoles of $[O H^{-}]$ in mixture $= (4.5 - 4) m m o l$
$= 0.5 m m o l$
Concentration of $O H^{-}$ , in mixture $= \frac{0.5}{40 + 10} M$
$= \frac{0.5}{50} M = 0.01 M$
$p O H = - log [O H^{-}] = - log [1 \times 10^{- 2}] = 2$
From, $p H + p O H = 14$
$p H = 14 - 2 = 12$ .

Suggest Corrections

Similar questions

p H

40 m L

0.1 M

H C l

10 m L

0.45 M

N a O H

10 m l

\frac{M}{200} H_{2} S O_{4}

40 m l

\frac{M}{200} H_{2} S O_{4}

p H

H_{2} S O_{4} .

N a C l

200 m l

N a O H

40 m l

H C l

Join BYJU'S Learning Program