Question

Calculate the pH of aqueous solution of 0.05 mol per decimeter cube of trioxonitrate(V) acid

Answer 1

First we need to translate the name to the compound.

trioxonitrate means we have 3 oxygens and one nigrogen.

The V means the N is 5+. If each O is 2-, we have a total of 5+ from the N and 6- from the 3 O, so the trioxonitrate(V) has a charge of 1-. Sicne it is acid, the charge is neutralized by H⁺, and since the charge is only 1-, we only need 1 H⁺.

So, the compound is HNO₃. This compound is a strong acid, which means that pratically all of the compounds dissociate in aqueous solution.

`HNO_3\to H^++NO^-_3_{}`

If we have a solution of 0.05 mol/dm³ and each dissociates, we will have the same concentration of H⁺:

`\lbrack H^+\rbrack=0.05mol\/dm^3`

And since dm³ is the same as L, we have:

`\lbrack H^+\rbrack=0.05mol\/L`

The pH can be calculated as:

`pH=-\log _(10)\lbrack H^+\rbrack`

Where [H⁺] is in mol/L, so:

`\begin{gathered} pH=-\log _(10)0.05 \\ pH=-(-1.301\ldots) \\ pH=1.301\ldots \\ pH\approx1.3 \end{gathered}`

So, the pH is approximately 1.3.