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PH of mixture of 500 ml 0.1 molar ch3cooh and 250 ml of 0.1 molar naoh solution is?
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PH of mixture of 500 ml 0.1 molar ch3cooh and 250 ml of 0.1 molar naoh...
Calculation of pH of Mixture of CH3COOH and NaOH Solution

The pH of a mixture of CH3COOH and NaOH solution can be calculated using the following steps:



  1. Write the balanced chemical equation for the reaction between CH3COOH and NaOH.

  2. Determine the moles of CH3COOH and NaOH in the solution.

  3. Identify the limiting reagent and calculate the amount of excess reagent remaining.

  4. Calculate the concentration of the resulting salt.

  5. Use the concentration of the salt to calculate the pH of the solution.



1. Balanced chemical equation

The balanced chemical equation for the reaction between CH3COOH and NaOH is:


CH3COOH + NaOH → CH3COONa + H2O


2. Moles of CH3COOH and NaOH

The moles of CH3COOH and NaOH in the solution can be calculated using the formula:


moles = concentration × volume


For CH3COOH:


moles CH3COOH = 0.1 mol/L × 0.5 L = 0.05 mol


For NaOH:


moles NaOH = 0.1 mol/L × 0.25 L = 0.025 mol


3. Limiting reagent and excess reagent

The limiting reagent is the reactant that is completely consumed in the reaction. In this case, NaOH is the limiting reagent because it has fewer moles than CH3COOH. The excess reagent is the reactant that remains after the limiting reagent is completely consumed. In this case, CH3COOH is the excess reagent.


The amount of excess CH3COOH remaining can be calculated using the formula:


moles of excess CH3COOH = moles CH3COOH - moles NaOH


moles of excess CH3COOH = 0.05 mol - 0.025 mol = 0.025 mol


4. Concentration of resulting salt

The concentration of the resulting salt, CH3COONa, can be calculated using the formula:


concentration of CH3COONa = moles of CH3COONa / total volume of solution


moles of CH3COONa = moles NaOH = 0.025 mol


total volume of solution = 0.5 L + 0.25 L = 0.75 L


concentration of CH3COONa = 0.025 mol / 0.75 L = 0.033 mol/L


5. pH of the solution

The pH of the solution can be calculated using the formula:


pH = pKa + log([A-] / [HA])


where:


pKa = 4
Community Answer
PH of mixture of 500 ml 0.1 molar ch3cooh and 250 ml of 0.1 molar naoh...
Here the H+ ions will be produced only by weak acid (CH3COOH)....
CH3COOH>>>>> CH3COO- + H+
0.1M. 0.1M. 0.1M
PH= -log [H+]
PH=-log(0.1)
PH=1
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PH of mixture of 500 ml 0.1 molar ch3cooh and 250 ml of 0.1 molar naoh solution is?
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