Question

a 27.1 ml sample of 0.348 m dimethylamine, (ch3)2nh, is titrated with 0.292 m hydrochloric acid. at the titration midpoint, the ph is
use the tables link in the references for any equilibrium constants that are required.
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Explanation:

Step1: Recall the concept of titration midpoint for a weak base

For the titration of a weak base (dimethylamine, \((CH_3)_2NH\)) with a strong acid (HCl), at the midpoint of the titration (half - equivalence point), the concentration of the weak base \([(CH_3)_2NH]\) is equal to the concentration of its conjugate acid \([(CH_3)_2NH_2^+]\). The relationship between \(pOH\) and \(pK_b\) at this point is given by the Henderson - Hasselbalch equation for bases: \(pOH=pK_b+\log\frac{[(CH_3)_2NH_2^+]}{[(CH_3)_2NH]}\). Since \([(CH_3)_2NH_2^+]=[(CH_3)_2NH]\) at the mid - point, \(\log\frac{[(CH_3)_2NH_2^+]}{[(CH_3)_2NH]} = \log(1)=0\), so \(pOH = pK_b\). Then we can find \(pH\) using the relationship \(pH + pOH=14\).

First, we need to find the \(K_b\) of dimethylamine. From reference tables, the \(K_b\) of \((CH_3)_2NH\) is \(5.4\times10^{-4}\). Then \(pK_b=-\log(K_b)=-\log(5.4\times10^{-4})\approx3.27\).

Step2: Calculate \(pOH\) at mid - point

As we established, at the mid - point of the titration of a weak base with a strong acid, \(pOH = pK_b\). So \(pOH\approx3.27\).

Step3: Calculate \(pH\) from \(pOH\)

Using the formula \(pH=14 - pOH\), we substitute \(pOH = 3.27\) into the formula: \(pH = 14-3.27 = 10.73\)

Answer:

\(10.73\)