Question

What is the $pH$ of a $1 \times {10^{ - 5}}M$ solution of sulfuric acid?

Answer 1

We have to know that, the $pH$ is a proportion of hydrogen particle focus, a proportion of the acidity or alkalinity of an answer. The $pH$ scale for the most part goes from zero to $14$. Watery arrangements at $25^\circ C$ with a $pH$ under seven are acidic, while those with a $pH$ more prominent than seven are fundamental or soluble.

Complete step by step answer:
We have to see that the sulphuric acid is diprotic, and it separates in two stages. The balanced chemical equation is given,
${H_2}S{O_4} \to {H^ + } + HSO_4^ -$
This has a harmony steady that is enormous and can be thought to be "absolute separation". The subsequent advance is the separation of the bi-sulphate particle $HSO_4^ -$, yet this just in part separates and the balance steady of this response is a lot of lower, truth be told the ${K_a}$ an incentive for bi-sulphate is just $0.0120$.
Then, bi-sulphate dissociates as,
$HSO_4^ - \to {H^ + } + SO_4^{2 - }$
Where,
${K_a} = \dfrac{{[{H^ + }][SO_4^{2 - }]}}{{[HSO_4^ - ]}}$
If we let the hydrogen ion concentration be $X$, then
$X = [{H^ + }] = [SO_4^{2 - }]$
Therefore,
$[{H^ + }][SO_4^{2 - }] = {[X]^2}$
The centralization of bisulphate staying in arrangement will be the first grouping of bisulphate less the sum that separated, which is $0.00001 - X$.
We have to know that,
${K_a}$= $0.0120$
Then,
$0.0120 = \dfrac{{{{\left[ X \right]}^2}}}{{\left( {0.00001 - X} \right)}}$
${[X]^2} = 0.0120 \times (0.00001 - X)$
${X^2} = 0.0000001 - 0.0120X$
Now, the above expression is rearranged,
${X^2} + 0.0120x - 0.0000001 = 0$ The other negative root is negligible.
Then, the above $X$ value is equal to $[{H^ + }]$, the value of $[{H^ + }]$ is 0.00001. Then, $(0.00000833 + 0.00001) = 0.00001833$
Finally,
$pH = - \log [{H^ + }]$
Now we can add the given value we get,
$pH = - \log 0.000001833$
On simplification we get,
$pH = 4.73$
Hence,
The value of $pH$ is $4.73$.

Note: We have to know $[{H^ + }]$ means the movement or viable grouping of ${H^ + }$ particles. However long the grouping of a acid (strong acid) is not huge, there is not a lot of contrast among $[{H^ + }]$ and a $[{H^ + }]$ yet at the constraints of the $pH$ range (characterized by the dissolvable), 'a' assumes a significant part.