Question

patrick makes his solution and standardizes it using khp (204.22 g/mol). he uses 0.7863 g khp dissolved in water and an indicator to identify the endpoint. the reaction requires 35.10 ml of the naoh solution. what is the actual concentration of the naoh solution? khp + naoh → nakp + h₂o(l) ? m naoh

Explanation:

Step1: Calculate moles of KHP

Moles of KHP = mass / molar mass. Mass of KHP is 0.7863 g, molar mass is 204.22 g/mol. So moles of KHP = $\frac{0.7863\ g}{204.22\ g/mol}$.
$\frac{0.7863}{204.22} \approx 0.00385\ mol$

Step2: Determine moles of NaOH

From the balanced equation, 1 mole of KHP reacts with 1 mole of NaOH. So moles of NaOH = moles of KHP = 0.00385 mol.

Step3: Calculate concentration of NaOH

Concentration (M) = moles / volume (in L). Volume of NaOH is 35.10 mL = 0.03510 L. So concentration = $\frac{0.00385\ mol}{0.03510\ L}$.
$\frac{0.00385}{0.03510} \approx 0.110\ M$

Answer:

0.110 (or more precisely, after detailed calculation: $\frac{0.7863}{204.22\times0.03510} \approx 0.110\ M$)