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6-2.Equilibrium-II (Ionic Equilibrium)
hard

$10^{-3}\, M\, H_2CO_3$ માટે જો = $10$$\%$ હોય તો $pH$ ના મુલ્યની ગણતરી શું હશે ?

A

$3.2$

B

$3.7$

C

$4.3$

D

$4.9$

Solution

સપ્રમાણતા = $M \times  V.F. = 10^{-3} \times  2 = 2 \times  10^{-3}$

$[H^+] = C,  \alpha = 2 \times 10^{-3} \times 0.1$.

$ = 2 \times {10^{ – 3}} \times \frac{1}{{10}} = 2 \times {10^{ – 4}}$

$= 2 \times 10^{-4}$

$pH = – log [H^+] = -log(2 \times 10^{-4})$

$= -log 2 – log 10^{-4} = -0.3010 + 4 = 3.7$

Standard 11
Chemistry

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