# 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?

Nov 12, 2015

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 ${H}_{3} {O}^{+}$.

Moles of nitric acid: $26.0 \times {10}^{- 3} \cdot L \times 8.00 \cdot m o l \cdot {L}^{-} 1 = 0.208 \cdot m o l$ $H N {O}_{3} \left(a q\right)$.

And, moles of hydrochloric acid: $88.0 \times {10}^{- 3} \cdot L \times 5.00 \cdot m o l \cdot {L}^{-} 1 = 0.440 \cdot m o l$ $H C l \left(a q\right)$.

This molar quantity is diluted to $1.00$ $L$. Concentration in moles/Litre = $\frac{\left(0.208 + 0.440\right) \cdot m o l}{1 L}$ $=$ $0.648 \cdot m o l \cdot {L}^{- 1} .$

Now we know that water undergoes autoprotolysis:
${H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}^{+} + O {H}^{-}$. This is another equilibrium reaction, and the ion product $\left[{H}^{+}\right] \left[O {H}^{-}\right]$ $=$ ${K}_{w}$. This constant, ${K}_{w}$ $=$ ${10}^{- 14}$ at $298 K$.

So $\left[{H}^{+}\right] = 0.648 \cdot m o l \cdot {L}^{- 1}$; $\left[O {H}^{-}\right] = {K}_{w} / \left[\left[{H}^{+}\right]\right]$ $=$ $\frac{{10}^{-} 14}{0.648}$ $=$ ??

$p H$ $=$ -log_(10)[H^+] = -log_(10)(0.648) = ??

Alternatively, we know further that $p H + p O H = 14$. Once you have $p H$, $p O H$ is easy to find. Take the antilogarithm of this to get $\left[O {H}^{-}\right]$.