Question

a chemistry student weighs out 0.193 g of citric acid ($h_3c_6h_5o_7$), a triprotic acid, into a 250 ml volumetric flask and dilutes to the mark with distilled water. he plans to titrate the acid with 0.1700m naoh solution. calculate the volume of naoh solution the student will need to add to reach the final equivalence point. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate moles of citric acid

The molar - mass of citric acid ($H_3C_6H_5O_7$) is:
$M=(3\times1.01 + 6\times12.01+5\times1.01 + 7\times16.00)\ g/mol=(3.03+72.06 + 5.05+112.00)\ g/mol = 192.14\ g/mol$.
The moles of citric acid, $n_{acid}=\frac{m}{M}$, where $m = 0.193\ g$ and $M = 192.14\ g/mol$.
$n_{acid}=\frac{0.193\ g}{192.14\ g/mol}=0.0010045\ mol$.

Step2: Determine the mole - ratio of acid and base

Citric acid ($H_3C_6H_5O_7$) is a triprotic acid, and the reaction with $NaOH$ is $H_3C_6H_5O_7+3NaOH
ightarrow Na_3C_6H_5O_7 + 3H_2O$. The mole - ratio of $H_3C_6H_5O_7$ to $NaOH$ is $n_{acid}:n_{base}=1:3$.
So the moles of $NaOH$ required, $n_{base}=3\times n_{acid}$.
$n_{base}=3\times0.0010045\ mol = 0.0030135\ mol$.

Step3: Calculate the volume of $NaOH$ solution

The molarity of $NaOH$ solution, $M_{base}=0.1700\ M$.
We know that $M=\frac{n}{V}$, so $V=\frac{n}{M}$.
$V_{base}=\frac{n_{base}}{M_{base}}=\frac{0.0030135\ mol}{0.1700\ mol/L}=0.017726\ L$.
Convert liters to milliliters: $V_{base}=0.017726\ L\times1000\ mL/L = 17.7\ mL$ (3 significant digits).

Answer:

$17.7\ mL$