Question

114.7 ml of 0.0371 m triprotic acid is neutralized using 24.1 ml of sodium hydroxide. what is the concentration (in mol/l) of the base? please report just the numerical answer to three significant figures and do not enter units.

Explanation:

Step1: Calculate moles of acid

The formula for moles \(n = C\times V\) (where \(C\) is concentration and \(V\) is volume in liters).
For the acid, \(V_{acid}=114.7\space mL = 0.1147\space L\) and \(C_{acid}=0.0371\space M\).
\(n_{acid}=C_{acid}\times V_{acid}=0.0371\times0.1147 = 0.00425537\space mol\).
Since it is a tri - protic acid, the number of moles of \(H^{+}\) ions \(n_{H^{+}} = 3\times n_{acid}=3\times0.00425537=0.01276611\space mol\).

Step2: Relate moles of \(H^{+}\) and \(OH^{-}\)

In neutralization, \(n_{H^{+}}=n_{OH^{-}}\).
For the base (\(NaOH\)), \(n_{OH^{-}}=n_{NaOH}\) (since \(NaOH\) dissociates as \(NaOH
ightarrow Na^{+}+OH^{-}\)).
The volume of the base \(V_{base}=24.1\space mL = 0.0241\space L\).
Using the formula \(C=\frac{n}{V}\), \(C_{base}=\frac{n_{NaOH}}{V_{base}}\).
Since \(n_{NaOH}=n_{OH^{-}} = 0.01276611\space mol\) and \(V_{base}=0.0241\space L\), \(C_{base}=\frac{0.01276611}{0.0241}\)

Step3: Calculate the concentration of the base

\(C_{base}=\frac{0.01276611}{0.0241}\approx0.530\)

Answer:

\(0.530\)