what is the ph of the mixture when 25.00 ml of 0.200 m koh has been added to 25.00 ml of 0.200 m acetic acid?

When a strong base like potasium hydroxide (KOH ) reacts with weak acid like acetic acid (CH₃COOH) form a salt ( CH₃COOK ) and water . This type of reaction is called Neutralization reaction.

CH₃COOH (aq) + KOH (aq) ------> CH₃COO⁻K⁺ (aq) + H₂O(aq)

Given that

Hen25 ml of potasium hydroxide( KOH) will react with 25ml of acetic acid (CH₃COOH) of both 0.200m give potasium acetate ( Ch₃COO⁻K⁺) 25ml of strength 0.200m. The unreaceted acetic acid will be present there.
Number of moles of KOH = 0.200×25 ×10⁻³

= 5 × 10⁻³mol

Number of moles of CH₃COO⁻K⁺= 0.200×25×10⁻³ = 5 × 10 ⁻³mol
Number of moles of CH₃COOH = 5 × 10⁻³ mol

Now , using the Henderson-Hasselbalch equation to find the p-H value

p-H = pKa + log ( salt or conjugate base / weak acid ) ------(1)

where Kₐ is dissociation constant

Since, Kₐ value for acetic acid is 3.8×10⁻⁵

=> pKₐ = 4.74

put the values in (1)

p-H = 4.74 + log ( 5/5) = 4.74 + 0 = 4.74

So, p-H value for solution is 4.74 ..

p-H value of substance or solution refered the nature of solution is acidic or basic .

To learn more about p-H value , refer :

https://brainly.com/question/26424076

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