Question

1. a solution of sodium hydroxide is standardized against potassium hydrogen phthalate. from the following data, calculate the molarity of naoh solution. mass of khp used: 0.4536 g buret reading before titration: 0.23 ml buret reading at the end point: 31.26 ml

Explanation:

Step1: Calculate the volume of NaOH used

The volume of NaOH used \(V = V_{final}-V_{initial}\).
\(V=(31.26 - 0.23)\text{ mL}=31.03\text{ mL}=31.03\times10^{- 3}\text{ L}\)

Step2: Calculate the moles of KHP

The molar mass of KHP (\(C_{8}H_{5}KO_{4}\)) is \(M=(8\times12.01 + 5\times1.01+39.10 + 4\times16.00)\text{ g/mol}=204.23\text{ g/mol}\)
The moles of KHP, \(n_{KHP}=\frac{m}{M}\), where \(m = 0.4536\text{ g}\)
\(n_{KHP}=\frac{0.4536\text{ g}}{204.23\text{ g/mol}}\approx2.221\times10^{-3}\text{ mol}\)

Step3: Use the stoichiometry of the reaction

The reaction between NaOH and KHP is \(NaOH + C_{8}H_{5}KO_{4}\to C_{8}H_{4}KNaO_{4}+H_{2}O\), and the mole ratio \(n_{NaOH}:n_{KHP}=1:1\)
So \(n_{NaOH}=n_{KHP} = 2.221\times10^{-3}\text{ mol}\)

Step4: Calculate the molarity of NaOH

Molarity \(M=\frac{n}{V}\), where \(n = n_{NaOH}\) and \(V\) is the volume of NaOH in liters
\(M=\frac{2.221\times 10^{-3}\text{ mol}}{31.03\times10^{-3}\text{ L}}\approx0.0716\text{ mol/L}\)

Answer:

The molarity of the NaOH solution is approximately \(0.0716\text{ M}\)