The pH of the solution after 22.0 ml of NaOH has been added to the acid will be 5.75.
Volume of CH₃CH₂COOH = 10.0 ml
Molarity of CH₃CH₂COOH = 0.75 M
Volume of NaOH = 22 ml
Molarity of NaOH = 0.30 M
pH os solution = ?
Write balanced chemical equation
CH₃CH₂COOH + NaOH → CH₃CH₂COONa + H₂O
Molar ratios are equal:
Convert mol to mmol;
mmol of CH₃CH₂COONa = M2 x V2
mmol of CH₃CH₂COONa = 0.30 mol x 22.0 ml
mmol of CH₃CH₂COONa = 6.6 mmol
mmol of CH₃CH₂COOH = M2 x V2
mmol of CH₃CH₂COOH = 0.75 mol x 10.0 ml
mmol of CH₃CH₂COOH = 0.9 mmol
Calculate the pH of the solution by using the Henderson equation:
pH = -log Ka + log [salt/acid]
pH = -log 1.3 x 10⁻⁵ + log (6.6 / 0.9)
pH = 5.75
Complete question:
A 10.0-ml sample of 0.75 m ch3ch2cooh is titrated with 0.30 m NaOH. what is the ph of the solution after 22.0 ml of NaOH has been added to the acid? ka = 1.3 x 10⁻⁵
You can also learn about Henderson equation from the following question:
https://brainly.com/question/13423434
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