What volume of $2.138 \cdot m o l \cdot L^{- 1}$ $2.138molL^-1$ $N a O H$ $NaOH$ is required to react with a $25.0 \cdot m L$ $25.0mL$ volume of $0.3057 \cdot m o l \cdot L^{- 1}$ $0.3057mol*L^-1$ solution of $potassium pthalate$ $"potassium pthalate"$ , $1, 2 - C_{6} H_{4} (C O_{2} H) (C O_{2}^{-} K^{+})$ $1,2-C_6H_4(CO_2H)(CO_2^(-)K^(+))$ ?

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What volume of $2.138 \cdot m o l \cdot L^{- 1}$ $2.138molL^-1$ $N a O H$ $NaOH$ is required to react with a $25.0 \cdot m L$ $25.0mL$ volume of $0.3057 \cdot m o l \cdot L^{- 1}$ $0.3057mol*L^-1$ solution of $potassium pthalate$ $"potassium pthalate"$ , $1, 2 - C_{6} H_{4} (C O_{2} H) (C O_{2}^{-} K^{+})$ $1,2-C_6H_4(CO_2H)(CO_2^(-)K^(+))$ ?

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Answer 1 · 2017-03-02T13:46:23

Approx. $3.6 \cdot mL$ $"3.6"*"mL"$ .

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!):

$NaOH(aq) + {1,2-C}_{6} H_{4} ({CO}_{2}^{-} K^{+}) {CO}_{2} H \to {1,2-C}_{6} H_{4} ({CO}_{2}^{-} K^{+}) ({CO}_{2}^{-} {Na}^{+}) + H_{2} O$ $"NaOH(aq)" + "1,2-C"_6"H"_4("CO"_2^(-)"K"^(+))"CO"_2"H"rarr"1,2-C"_6"H"_4("CO"_2^(-)"K"^(+))("CO"_2^(-)"Na"^(+)) +"H"_2"O"$

There is thus $1:1 equivalence$ $"1:1 equivalence"$ between $moles of sodium hydroxide$ $"moles of sodium hydroxide"$ , and $moles of KHP.$ $"moles of KHP."$

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

$Moles of potassium bipthalate,$ $"Moles of potassium bipthalate,"$ $0.3057*mol*cancel(L^-1)xx25.0*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)$
$=7.643xx10^-3*mol$ .

And thus volume of standardized $"NaOH(aq)"$ 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 $0.2138*mol*L^-1$ $"NaOH(aq)"$ instead of $2.138*mol*L^-1$ $"NaOH(aq)"$ for this titration? Why, or why not?

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