# What is the pH of a .043 M HCl solution?

Apr 21, 2018

$- l g \left({H}_{C a t i o n}\right)$

#### Explanation:

pH = $- l g \left(0.043\right)$ = 1.34

pH is 1.34.

Apr 21, 2018

Approximately $1.37$.

#### Explanation:

The $\text{pH}$ of a solution is given by:

$\text{pH} = - \log \left[{H}^{+}\right]$

• $\left[{H}^{+}\right]$ is the hydrogen ion concentration in terms of molarity

The dissociation of $H C l$ is:

$H C l \left(a q\right) \to {H}^{+} \left(a q\right) + C {l}^{-} \left(a q\right)$

So, one mole of hydrochloric acid contains one mole of hydrogen ions. So here, there are $0.43 \setminus \text{mol/L}$ of hydrogen ions.

Therefore, its $\text{pH}$ will be:

$\text{pH} = - \log \left[0.043\right]$

$\approx 1.37$

Apr 21, 2018

Well, I make it $1.37$...bowling another googly...

#### Explanation:

By definition, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$...i.e. $\text{pouvoir hydrogène}$ for $\left[H C l\right]$ of $0.043 \cdot m o l \cdot {L}^{-} 1$ concentration...is given by....

$p H \equiv - {\log}_{10} \left(\left[H C l\right]\right) = - {\log}_{10} \left(0.043\right) = - \left(- 1.366\right) = 1.37$...

And we assume here, quite reasonably, that the strong acid $H C l$ undergoes complete protonolysis in the water solvent according to the reaction....

$H C l \left(a q\right) + {H}_{2} O \left(l\right) \rightarrow {\underbrace{{H}_{3} {O}^{+}}}_{\text{molar concentration of 0.043 mol/L}} + C {l}^{-}$

And so we take the logarithm of the initial $\left[H C l\right]$ concentration DIRECTLY....