Question

missed this? watch |we: acid-base titration; read section 5.7. you can click on the review link to access the section in your etext. a 16.00 ml sample of an unknown hclo4 solution requires titration with 18.38 ml of 0.1800 m naoh to reach the equivalence point. part a what is the concentration of the unknown hclo4 solution? the neutralization reaction is hclo4(aq) + naoh(aq) → h2o(l) + naclo4(aq) express the molarity to four significant figures. view available hint(s) m(hclo4) = m

Explanation:

Step1: Use the formula for acid - base titration at equivalence point

At the equivalence point of an acid - base titration \(n(\text{acid})=n(\text{base})\). For a monoprotic acid (\(\text{HClO}_4\)) and a monoprotic base (\(\text{NaOH}\)), \(M(\text{HClO}_4)\times V(\text{HClO}_4)=M(\text{NaOH})\times V(\text{NaOH})\).

Step2: Rearrange the formula to solve for \(M(\text{HClO}_4)\)

\(M(\text{HClO}_4)=\frac{M(\text{NaOH})\times V(\text{NaOH})}{V(\text{HClO}_4)}\)
Given \(M(\text{NaOH}) = 0.1800\space M\), \(V(\text{NaOH})=18.38\space mL\), and \(V(\text{HClO}_4)=16.00\space mL\)
Substitute the values into the formula:
\(M(\text{HClO}_4)=\frac{0.1800\space M\times18.38\space mL}{16.00\space mL}\)

Step3: Calculate the value

\(M(\text{HClO}_4)=\frac{0.1800\times18.38}{16.00}\space M\)
\(0.1800\times18.38 = 3.3084\)
\(\frac{3.3084}{16.00}= 0.206775\space M\)

Answer:

\(0.2068\space M\)