The value of the pH when 20.0 of 0.150 M KOH is mixed with 29.0ml of 0.300M HBr is 8.32
The chemical reaction between KOH and HBr is a neutralization reaction.
[tex]\mathbf{KOH + HBr \to KBr + H_2O}[/tex]
From the given reaction;
The number of moles of KOH = 20.0 × 0.150 M
= 3 mmol
The number of moles of HBr = 0.300 M × 29 ml
= 8.7 mmol
If 3mmol of KOH reacts with 8.7 mmol of HBr to give 3 mmol of KBr;
Then, the remaining amount of HBr is = (8.7 - 3.0) mmol
= 5.7 mmol
Now, using the Henderson Hasselbalch equation to determine the pH of the system, we have:
[tex]\mathbf{pH = pka + log \dfrac{[KBr] }{[HBr]}}[/tex]
[tex]\mathbf{pH = -log (2.5 \times 10^{-9})+ log( \dfrac{3 }{5.7})}[/tex]
[tex]\mathbf{pH =8.6021+(-0.27875)}[/tex]
pH ≅ 8.32
Learn more about the Henderson Hasselbalch equation here:
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