10. the concentration of oxalate ion ($c_2o_4^{2 - }$) in a sample can be determined by titration with a…

10. 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) 0.1821 m

10. 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) 0.1821 m

Answer

Explanation:

Step1: Determine moles of $MnO_4^-$

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

Step2: Use mole - ratio to find moles of $C_2O_4^{2 -}$

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

Step3: Calculate molarity of $C_2O_4^{2 -}$

$V_{C_2O_4^{2 -}} = 30.00\ mL=0.03000\ L$ $M_{C_2O_4^{2 -}}=\frac{n_{C_2O_4^{2 -}}}{V_{C_2O_4^{2 -}}}=\frac{0.0094598125\ mol}{0.03000\ L}=0.315327\ M\approx0.3152\ M$

Answer:

C. $0.3152\ M$