What is the ph of a solution which is 0.025 m in weak base and 0.037 m in the conjugate weak acid (ka = 7.1 × 10–6)?

Hello!

This question can be solved using the Henderson-Hasselbach equation, but first we need to know the value of pKa

pKa= -log(Ka)= -log(7,1*10⁻⁶)=5,15

Now, we apply the equation:

[tex]pH=pKa+log \frac{[A^{-}] }{[HA]} =5,15+log( \frac{0,025M}{0,037M} )=4,98[/tex]

So, the pH of this solution would be 4,98

Have a nice day!

Answer Link

aliasger2709 aliasger2709 · Answer 2 · 2022-02-28T07:26:34+01:00

pH is the concentration of the hydrogen and the hydroxide ion concentration in the solution and depicts the acidity and alkalinity. The pH of the solution is 4.68.

What are weak bases and conjugate weak acids?

Weak bases are the solutions of substances that do not dissociate completely in water. Conjugate acids are the product formed by the base and have a difference of one proton.

pKa can be calculated as,

[tex]\begin{aligned} \rm pKa & = \rm -log(Ka)\\\\& = \rm -log(7.1 \times 10^{-6})\\\\& =5.15 \end{aligned}[/tex]

The Henderson-Hasselbach equation is shown as,

[tex]\begin{aligned} \rm pH &= \rm pKa + log \dfrac{[A^{-}]}{[HA]}\\\\&= \rm 5.15 + log (\dfrac{0.025}{0.037})\\\\&= 4.98 \end{aligned}[/tex]

Therefore, 4.98 is the pH of the solution.

Learn more about the Henderson-Hasselbach equation here:

https://brainly.com/question/13651361