Question

1. using the actual molarity of the naoh solution, calculate the molarity of 26.00 ml of a h₃po₄ solution that requires 32.50 ml of naoh to reach the endpoint.

Explanation:

1. First, write the balanced chemical - equation for the reaction between \(H_3PO_4\) and \(NaOH\):

• The reaction is \(H_3PO_4 + 3NaOH=Na_3PO_4 + 3H_2O\).
• From the balanced equation, the mole - ratio of \(H_3PO_4\) to \(NaOH\) is \(n_{H_3PO_4}:n_{NaOH}=1:3\), or \(n_{H_3PO_4}=\frac{1}{3}n_{NaOH}\).

1. Recall the formula for the number of moles (\(n\)) in terms of molarity (\(M\)) and volume (\(V\)):

• The formula is \(n = M\times V\) (where \(V\) is in liters). Let \(M_{NaOH}\) be the molarity of the \(NaOH\) solution, \(V_{NaOH}=32.50\ mL = 0.03250\ L\), \(V_{H_3PO_4}=26.00\ mL = 0.02600\ L\), and \(M_{H_3PO_4}\) be the molarity of the \(H_3PO_4\) solution we want to find.
• We know that \(n_{H_3PO_4}=\frac{1}{3}n_{NaOH}\). Substituting \(n = M\times V\) into the mole - ratio equation, we get \(M_{H_3PO_4}\times V_{H_3PO_4}=\frac{1}{3}M_{NaOH}\times V_{NaOH}\).

1. Solve for \(M_{H_3PO_4}\):

• Rearrange the equation \(M_{H_3PO_4}\times V_{H_3PO_4}=\frac{1}{3}M_{NaOH}\times V_{NaOH}\) to isolate \(M_{H_3PO_4}\).
• \(M_{H_3PO_4}=\frac{M_{NaOH}\times V_{NaOH}}{3\times V_{H_3PO_4}}\).
• Substitute \(V_{NaOH}=0.03250\ L\) and \(V_{H_3PO_4}=0.02600\ L\) into the equation: \(M_{H_3PO_4}=\frac{M_{NaOH}\times0.03250}{3\times0.02600}\).
• First, calculate \(\frac{0.03250}{3\times0.02600}=\frac{0.03250}{0.07800}\approx0.4167\).
• So \(M_{H_3PO_4}=0.4167M_{NaOH}\).

Answer:

\(M_{H_3PO_4}=0.4167M_{NaOH}\)