# Question 8dd75

Mar 23, 2015

This is dilution. Apr 7, 2015

Rabigh's answer is very good, I just want to add an alternative approach to using the dilution calculations equation, ${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$.

You're dealing with ${\text{100 cm}}^{3}$ of a 0.1-M $N a O H$ solution. Because sodium hydroxide is a strong base, it will dissociate completely in aqueous solution to give $N {a}^{+}$ cations and $O {H}^{-}$ anions.

$N a O {H}_{\left(a q\right)} \to N {a}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

You can use this solution's molarity to determine the number of moles of sodium hydroxide, which is equal to the number of moles of $O {H}^{-}$ ions, you have in solution before adding water

$C = \frac{n}{V} \implies n = C \cdot V$

n_("NaOH") = "0.1 M" * 100 * 10^(-3)"L" = "0.01 moles NaOH"

When you add the ${\text{100 cm}}^{3}$ of water, the number of moles of sodium hydroxide remains unchanged; the only thing that changes is the volume of the solution, which implies that the concentration of $O {H}^{-}$ will change as well.

Think of it like this: same number of moles, twice the volume $\to$ half the initial concentration

C_("new") = n/V_("total") = "0.01 moles"/((100 + 100) * 10^(-3)"L") = "0.05 M"#

Use this concentration to determine pOH

$p O H = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left(0.05\right) = 1.3$

Therefore, pH is equal to

$p {H}_{\text{sol}} = 14 - p O H = 14 - 1.3 = \textcolor{g r e e n}{12.7}$