# What volume of 2.138*mol*L^-1 NaOH is required to react with a 25.0*mL volume of 0.3057*mol*L^-1 solution of "potassium pthalate", 1,2-C_6H_4(CO_2H)(CO_2^(-)K^(+))?

Mar 2, 2017

Approx. $\text{3.6"*"mL}$.

#### Explanation:

We need (i) a stoichiometrically balanced equation to represent the acid-base reaction (this should always be your priority in these types of questions!):

$\text{NaOH(aq)" + "1,2-C"_6"H"_4("CO"_2^(-)"K"^(+))"CO"_2"H"rarr"1,2-C"_6"H"_4("CO"_2^(-)"K"^(+))("CO"_2^(-)"Na"^(+)) +"H"_2"O}$

There is thus $\text{1:1 equivalence}$ between $\text{moles of sodium hydroxide}$, and $\text{moles of KHP.}$

And now (ii) we need to find the equivalent quantites of each reagent:

$\text{Moles of potassium bipthalate,}$ $0.3057 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 25.0 \cdot \cancel{m L} \times {10}^{-} 3 \cdot \cancel{L} \cdot \cancel{m {L}^{-} 1}$
$= 7.643 \times {10}^{-} 3 \cdot m o l$.

And thus volume of standardized $\text{NaOH(aq)}$ required is:

(7.643xx10^-3*cancel"mol")/(2.138*cancel"mol"*cancel("L"^-1))xx10^3*mL*cancel("L"^-1)=3.58*"mL"

Do you think it would have been better to have used $0.2138 \cdot m o l \cdot {L}^{-} 1$ $\text{NaOH(aq)}$ instead of $2.138 \cdot m o l \cdot {L}^{-} 1$ $\text{NaOH(aq)}$ for this titration? Why, or why not?