Question

1. weigh 0.5208 g of the primary standard khc₈h₄o₄ to standardize the naoh solution. at the stoichiometric point, 25.20 ml of the naoh solution is consumed. calculate the concentration of the naoh solution. m(khc₈h₄o₄) = 204.22 g·mol⁻¹ the mole ratio of khc₈h₄o₄ to naoh in the reaction is 1:1.

Explanation:

Step1: Calculate moles of $KHC_8H_4O_4$

$n_{KHC_8H_4O_4}=\frac{m}{M}=\frac{0.5208\ g}{204.22\ g\cdot mol^{-1}}$

Step2: Determine moles of $NaOH$

Since the mole - ratio of $KHC_8H_4O_4$ to $NaOH$ is 1:1, $n_{NaOH}=n_{KHC_8H_4O_4}=\frac{0.5208\ g}{204.22\ g\cdot mol^{-1}}$

Step3: Calculate concentration of $NaOH$

$c_{NaOH}=\frac{n_{NaOH}}{V_{NaOH}}$, where $V_{NaOH}=25.20\ mL = 0.02520\ L$
$c_{NaOH}=\frac{\frac{0.5208\ g}{204.22\ g\cdot mol^{-1}}}{0.02520\ L}$
$c_{NaOH}=\frac{0.5208}{204.22\times0.02520}\ mol\cdot L^{-1}\approx0.1000\ mol\cdot L^{-1}$

Answer:

$0.1000\ mol\cdot L^{-1}$