Question

exercise 5.66 - enhanced - with expanded hints
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 ( mathrm{h}_{3} mathrm{po}_{4} ) solution is titrated with a ( 0.100 mathrm{m} mathrm{naoh} ) solution. the equivalence point is reached when ( 26.58 mathrm{~ml} ) of
naoh solution is added.
part a
the neutralization reaction is
mathrm{h}_{3} mathrm{po}_{4}(mathrm{aq})+3 mathrm{naoh}(mathrm{aq}) \to 3 mathrm{h}_{2} mathrm{o}(mathrm{l})+mathrm{na}_{3} mathrm{po}_{4}(mathrm{aq})
what is the concentration of the unknown ( mathrm{h}_{3} mathrm{po}_{4} ) solution?
express the molarity to three significant figures.
view available hint(s)
concentration ( = )

Explanation:

Step1: Calculate the moles of NaOH

Use the formula \(n = M\times V\). Here, \(M = 0.100\space M\) and \(V=26.58\space mL=26.58\times10^{- 3}\space L\).
\(n_{NaOH}=0.100\space M\times26.58\times 10^{-3}\space L = 2.658\times10^{-3}\space mol\)

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 molarity of \(H_3PO_4\)

Use the formula \(M=\frac{n}{V}\). Here, \(n = 8.86\times10^{-4}\space mol\) and \(V = 35.00\space mL=35.00\times10^{-3}\space L\)
\(M_{H_3PO_4}=\frac{8.86\times10^{-4}\space mol}{35.00\times10^{-3}\space L}=0.0253\space M\)

Answer:

\(0.0253\space M\)