Question

1. the concentration of oxalate ion ($c_2o_4^{2 - }$) in a sample can be determined by titration with a solution of permanganate ion ($mno_4^-$) of known concentration. the balanced net - ionic equation for this reaction is $2mno_4^-+5c_2o_4^{2 - }+16h^+\to2mn^{2 + }+8h_2o + 10co_2$. a 30.00 ml sample of an oxalate solution is found to react completely with 21.93 ml of a 0.1725 m solution of $mno_4^-$. what is the oxalate ion concentration (in molarity, m) in the sample? a) 0.4312 m b) 0.05044 m c) 0.3152 m d) 0.02914 m e) 0.02814 m

Explanation:

Step1: Identify the mole - ratio

From the balanced equation $2MnO_4^-+5C_2O_4^{2 -}+16H^+\to2Mn^{2 +}+8H_2O + 10CO_2$, the mole - ratio of $MnO_4^-$ to $C_2O_4^{2 -}$ is $n_{MnO_4^-}:n_{C_2O_4^{2 -}}=2:5$.

Step2: Calculate the moles of $MnO_4^-$

Use the formula $n = M\times V$, where $M$ is the molarity and $V$ is the volume in liters. Given $M_{MnO_4^-}=0.1725\ M$ and $V_{MnO_4^-}=21.93\ mL = 0.02193\ L$. So, $n_{MnO_4^-}=M_{MnO_4^-}\times V_{MnO_4^-}=0.1725\ mol/L\times0.02193\ L = 0.003783925\ mol$.

Step3: Calculate the moles of $C_2O_4^{2 -}$

Since $\frac{n_{C_2O_4^{2 -}}}{n_{MnO_4^-}}=\frac{5}{2}$, then $n_{C_2O_4^{2 -}}=\frac{5}{2}\times n_{MnO_4^-}=\frac{5}{2}\times0.003783925\ mol = 0.0094598125\ mol$.

Step4: Calculate the molarity of $C_2O_4^{2 -}$

The volume of the $C_2O_4^{2 -}$ solution is $V_{C_2O_4^{2 -}}=30.00\ mL = 0.03000\ L$. Using the formula $M=\frac{n}{V}$, we have $M_{C_2O_4^{2 -}}=\frac{n_{C_2O_4^{2 -}}}{V_{C_2O_4^{2 -}}}=\frac{0.0094598125\ mol}{0.03000\ L}=0.3153\ M\approx0.3152\ M$.

Answer:

C. $0.3152\ M$