Question

missed this? watch iwe: acid-base titration; read section 5.7. you can click on the review link to access the section in your etext. a 35.00 ml sample of an unknown h₃po₄ solution is titrated with a 0.100 m naoh solution. the equivalence point is reached when 26.58 ml of naoh solution is added. part a the neutralization reaction is h₃po₄(aq) + 3naoh(aq) → 3h₂o(l) + na₃po₄(aq) what is the concentration of the unknown h₃po₄ solution? express the molarity to three significant figures. view available hint(s) concentration = m

Explanation:

Step1: Calculate the moles of \(NaOH\)

Use the formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, \(V\) is volume in liters).
The volume of \(NaOH\), \(V_{NaOH}=26.58\space mL = 26.58\times10^{- 3}\space L\), and \(C_{NaOH}=0.100\space M\).
\(n_{NaOH}=C_{NaOH}\times V_{NaOH}=0.100\space mol/L\times26.58\times 10^{-3}\space L = 2.658\times10^{-3}\space mol\)

Step2: Relate moles of \(NaOH\) to moles of \(H_3PO_4\)

From the balanced equation \(H_{3}PO_{4}(aq)+3NaOH(aq)\to3H_{2}O(l)+Na_{3}PO_{4}(aq)\), the mole ratio \(n_{H_3PO_4}:n_{NaOH}=1:3\).
So \(n_{H_3PO_4}=\frac{n_{NaOH}}{3}\)
\(n_{H_3PO_4}=\frac{2.658\times 10^{-3}\space mol}{3}=8.86\times10^{-4}\space mol\)

Step3: Calculate the concentration of \(H_3PO_4\)

The volume of \(H_3PO_4\), \(V_{H_3PO_4}=35.00\space mL=35.00\times10^{-3}\space L\)
Use the formula \(C=\frac{n}{V}\) (where \(C\) is concentration, \(n\) is moles, \(V\) is volume in liters)
\(C_{H_3PO_4}=\frac{n_{H_3PO_4}}{V_{H_3PO_4}}=\frac{8.86\times 10^{-4}\space mol}{35.00\times10^{-3}\space L}\)
\(C_{H_3PO_4}=0.0253\space M\)

Answer:

\(0.0253\space M\)