# Some help please question is below as a image thank you? "HNO"_3 is a strong acid. [What is] the "pH" of a 0.54color(white)(l)"mol"*"dm"^(-3) solution?

May 25, 2018

$\text{pH} \approx 0.62$

#### Explanation:

The question states that nitric acid ${\text{HNO}}_{3}$ is a strong acid, meaning that it would completely disassociate to produce hydronium ions and conjugate base anions when dissolved in water. Therefore the process

${\text{HNO"_3(aq) color(purple)(to) "H"^(+)(aq) + "NO}}_{3}^{-} \left(a q\right)$ or equivalently
${\text{HNO"_3(aq) + "H"_2"O"(l) color(purple)(to) "H"_3"O"^(+)(aq)+"NO}}_{3}^{-} \left(a q\right)$

would proceed to completion, such that

$n \left({\text{H"_3"O"^(+)(aq))=n("H"^(+)(aq))=n("HNO}}_{3} \left(a q\right)\right)$

$c = \frac{n}{V}$,

Assuming that the volume of the solution remains constant during this process, the stochiochemical relationship between the number of moles of reactants and products would also hold for their respective concentrations.

["H"_3"O"^(+)(aq)]=["H"^(+)(aq)]=["HNO"_3(aq)]=0.54color(white)(l)"mol"*"dm"^(-3)

Hence by definition,

"pH"=-log["H"^(+)]=-log(0.54)~~0.62

Note that the number of decimal places in the $\text{pH}$ measurement shall reflect the number of significant figures in the $\left[{\text{H}}^{+}\right]$ measurement. The $\left[{\text{HNO}}_{3}\right]$ and hence $\left[{\text{H}}^{+}\right]$ measurements in this question are accurate to two significant figures; therefore the calculated $\text{pH}$ value shall be rounded to two decimal places.