How many $mL$ $"mL"$ of $H_{2} C_{2} O_{4}$ $"H"_2"C"_2"O"_4$ can be neutralized by $23.65 mL$ $"23.65 mL"$ of $1.115 M$ $"1.115 M"$ $NaOH$ $"NaOH"$ ?

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How many $mL$ $"mL"$ of $H_{2} C_{2} O_{4}$ $"H"_2"C"_2"O"_4$ can be neutralized by $23.65 mL$ $"23.65 mL"$ of $1.115 M$ $"1.115 M"$ $NaOH$ $"NaOH"$ ?

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Answer 1 · 2014-12-05T22:59:08

The volume of oxalic acid required is $6.75 m L$ $6.75 mL$ .

First, start with the balanced chemical equation for this particular neutralization reaction

$H_{2} C_{2} O_{4} (a q) + 2 N a O H (a q) \to N a_{2} C_{2} O_{4} (a q) + 2 H_{2} O (l)$ $H_2C_2O_4(aq) + 2NaOH(aq) -> Na_2C_2O_4(aq) + 2H_2O(l)$

From the chemical reaction we can see that $1$ $1$ mole of $H_{2} C_{2} O_{4}$ $H_2C_2O_4$ needs $2$ $2$ moles of $N a O H$ $NaOH$ in order to produce $1$ $1$ mole of $N a C_{2} O_{4}$ $NaC_2O_4$ and $2$ $2$ moles of $H_{2} O$ $H_2O$ .

We can determine the number of moles of $N a O H$ $NaOH$ by using

$n_{N a O H} = C \cdot V = 1.115 \frac{m o l e s}{L} \cdot 0.02365 L = 0.026 m o l e s$ $n_(NaOH) = C * V = 1.115 (mol es)/(L) *0.02365 L = 0.026 mo l es$

This means that we need

$n_{H_{2} C_{2} O_{4}} = \frac{n_{N a O H}}{2} = \frac{0.026}{2} = 0.013 m o l e s$ $n_(H_2C_2O_4) = (n_(NaOH))/2 = 0.026/2 = 0.013 m o l es$ of oxalic acid.

Therefore, the volume of oxalic acid can be calculated from

$C_{H_{2} C_{2} O_{4}} = \frac{n_{H_{2} C_{2} O_{4}}}{V_{H_{2} C_{2} O_{4}}}$ $C_(H_2C_2O_4) = n_(H_2C_2O_4)/V_(H_2C_2O_4)$

$V_{H_{2} C_{2} O_{4}} = \frac{0.013 m o l e s}{1.978 \frac{m o l e s}{L}} = 6.57 m L$ $V_(H_2C_2O_4) = (0.013 mo l es)/(1.978 (mo l es)/L) = 6.57 mL$

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How many $mL$ $"mL"$ of $H_{2} C_{2} O_{4}$ $"H"_2"C"_2"O"_4$ can be neutralized by $23.65 mL$ $"23.65 mL"$ of $1.115 M$ $"1.115 M"$ $NaOH$ $"NaOH"$ ?

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How many $mL$ $"mL"$ of $H_{2} C_{2} O_{4}$ $"H"_2"C"_2"O"_4$ can be neutralized by $23.65 mL$ $"23.65 mL"$ of $1.115 M$ $"1.115 M"$ $NaOH$ $"NaOH"$ ?