Question

Suppose you had to prepare some NaOH solution according to the directions given in Part A. However, while using a 50-mL graduated cylinder to add the distilled water, you miscounted and only added 150 mL instead of 200 mL. How will this mistake influence the Ka value determined in Part B?

Answer 1

`\begin{array}{l}{\text { The mistake of adding less amount of as assigned distilled water will not affect the acid }} \\ {\text { dissociation constant value`

Step-by-step explanation:

The quantitative measurement of strength of an acid in solution named as acid dissociation constant
`^(\prime \prime) \mathrm{K}_(2)^(\prime \prime)`. For acid-base titration this analysis is preferred and it is equilibrium constant for a chemical reaction.
`\mathrm{k}_{\mathrm{a}}`is constant despite of concentration, for example it measures breakdown of an acid into the conjugate base
`\mathrm{A}^(-)`and hydrogen ion
`\mathrm{H}^(+) \text {in following equation: }`

`\mathrm{HA} \leftrightarrow \mathrm{H}^(+)+\mathrm{A}^(-)`

`\underline{\mathbf{k}}_{\mathbf{a}}=\frac{\left[\mathrm{H}^(+)\right] \cdot[\mathrm{A}]}{[\mathrm{HA}]}`

Here amount of
`\mathrm{H}^(+)`produced is proportional to the amount of H-A from which started. Therefore amount of
`\mathbf{k}_{\mathbf{a}}`remain constant for a particular acid despite a change in concentration of both acid or base with which it is reacting.