Ph of mixture of 500ml 0.1M ch3cooh and 250ml 0.1M Naoh Solution is (g...
Solution:
Given,
Volume of CH3COOH solution = 500 mL
Concentration of CH3COOH solution = 0.1 M
Volume of NaOH solution = 250 mL
Concentration of NaOH solution = 0.1 M
pKa of acetic acid = 4.74
Step 1: Write the balanced chemical equation for the reaction between CH3COOH and NaOH.
CH3COOH + NaOH → NaCH3COO + H2O
Step 2: Calculate the number of moles of CH3COOH and NaOH.
Number of moles of CH3COOH = Molarity × Volume (in liters) = 0.1 × 0.5 = 0.05 moles
Number of moles of NaOH = Molarity × Volume (in liters) = 0.1 × 0.25 = 0.025 moles
Step 3: Determine the limiting reagent.
Since NaOH is present in a lesser amount than CH3COOH, it is the limiting reagent.
Step 4: Calculate the number of moles of NaCH3COO formed.
From the balanced chemical equation, the stoichiometric ratio between NaOH and NaCH3COO is 1:1. Therefore, the number of moles of NaCH3COO formed is also 0.025 moles.
Step 5: Calculate the concentration of the resulting solution.
Total volume of the resulting solution = Volume of CH3COOH solution + Volume of NaOH solution = 500 mL + 250 mL = 750 mL = 0.75 L
Concentration of NaCH3COO = Number of moles of NaCH3COO / Total volume of the resulting solution = 0.025 / 0.75 = 0.033 M
Step 6: Calculate the pH of the resulting solution.
Using the Henderson-Hasselbalch equation,
pH = pKa + log ([A-]/[HA])
where [A-] is the concentration of NaCH3COO and [HA] is the concentration of CH3COOH.
Substituting the values,
pH = 4.74 + log (0.033/0.05)
pH = 4.74 - 0.23
pH = 4.51
Therefore, the pH of the resulting solution is 4.51.
Conclusion:
The pH of the mixture of 500 mL 0.1 M CH3COOH and 250 mL 0.1 M NaOH solution is 4.51.
Ph of mixture of 500ml 0.1M ch3cooh and 250ml 0.1M Naoh Solution is (g...
PH will be same 4.74 . if yes then ask me solution of this on telegram here I am unable to send. my I'd
https://t.me/rajeev1029kumar
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