Question

A buffer containing 0.2000 m of acid, ha, and 0.1500 m of its conjugate base, a−, has a ph of 3.35. what is the ph after 0.0015 mol of naoh is added to 0.5000 l of this solution?

Answer 1

After 0.0015 mol of NAOH is added to 0.5000 l of this solution, the pH is 3.37.

Steps

First, modify the Henderson-Hassle-batch equation to find the acid's pka.

$pH=pka + log\frac{[A]}{[HA]}$

$pka = pH-log\frac{[A]}{[HA]}$

$=3.35- log \frac{[0.1500]}{[ 0.2000 ]}$

$pka =3.47$

Molarity of the added NaOH.

$M=\frac{mol}{L}$

$=\frac{0.0015 mol}{ 0.5000L}$

= 0.003 M

In water, NaOH will dissociate

$NaOH- > Na^{+} + OH^{-}$

strong base OH will react with HA

HA+ OH --> $H_{2} O$ + A

the reaction also produced A. as, all OH are consumed in the reaction

HA will be decreased by 0.003 M and A will be increased by 0.003 M.

$HA= 0.2000-0.003 \\= 0.1970$

$A=0.1500+0.003=0.1530 M$

solving new pH using pka and the new values of [HA] and [A]

$pH=pka + log\frac{[A]}{[HA]}$

$=3.35 + log\frac{[0.1530]}{[0.1970]}$

pH=3.37 is the ph after 0.0015 mol of naoh is added to 0.5000 l of this solution

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