Question

pH of a dibasic acid is 3.699. it's molarity is​

Answer 1

Hello!!

pH = -logH+

3.699 = -log H

(taking log both side!)

10 - 3.699 = H

1.99986 x 10-⁴ = H

for dibasic acid it'll become half!!

Answer 2

The molarity of dibasic acid is $9.95\times 10^{-5}M$

Explanation:

To calculate the concentration of dibasic acid, we need to calculate teh concentration of hydrogen ion first. To calculate the hydrogen ion concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

pH = 3.699

Putting values in above equation, we get:

$3.699=-\log[H^+]$

$[H^+]=10^{-3.699}=1.99\times 10^{-4}M$

As, dibasic acid produces 2 moles of hydrogen ion. So, the concentration of acid will be half the concentration of hydrogen ion.

Concentration of dibasic acid = $\frac{[H^+]}{2}=\frac{1.99\times 10^{-4}}{2}=9.95\times 10^{-5}M$

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