Question

A solution is prepared by mixing 88.0 mL of 5.00 M HCl and 26.0 mL of 8.00 M HNO3. Water is then added until the final volume is 1.00 L. How would you calculate [H+], [OH -], and the pH for this solution? please explain?

Answer 1

This is a typical acid/base equilibrium problem, that involves the use of logarithms.

Explanation:

We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric `H_3O^+`.

Moles of nitric acid: `26.0xx10^(-3)*L xx 8.00*mol*L^-1 = 0.208*mol` `HNO_3(aq)`.

And, moles of hydrochloric acid: `88.0xx10^(-3)*L xx 5.00*mol*L^-1 = 0.440*mol` `HCl(aq)`.

This molar quantity is diluted to `1.00` `L`. Concentration in moles/Litre = `((0.208+0.440)* mol)/(1L)` `=` `0.648*mol*L^(-1).`

Now we know that water undergoes autoprotolysis:
`H_2O(l) rightleftharpoons H^+ +OH^-`. This is another equilibrium reaction, and the ion product `[H^+][OH^-]` `=` `K_w`. This constant, `K_w` `=` `10^(-14)` at `298K`.

So `[H^+] =0.648*mol*L^(-1)`; `[OH^-] = K_w/[[H^+]]` `=` `(10^-14)/0.648` `=` `??`

`pH` `=` `-log_(10)[H^+] = -log_(10)(0.648) = ??`

Alternatively, we know further that `pH + pOH =14`. Once you have `pH`, `pOH` is easy to find. Take the antilogarithm of this to get `[OH^-]`.