Calculate the pH of a solution obtained by mixing 2 ml of HCl of pH 2 and 3ml of solution of KOH pH 12?

Question

Answer

Calculating the pH of a Solution Obtained by Mixing HCl and KOH

Step 1: Calculate the moles of HCl

Given that 2 ml of HCl solution of pH 2 is added to the mixture, we need to first calculate the moles of HCl present in the solution.

pH = -log[H+]

2 = -log[H+]

[H+] = 10^-2 M

Number of moles of HCl = Molarity x Volume(in L) = 10^-2 x 2/1000 = 2 x 10^-5 moles

Step 2: Calculate the moles of KOH

Similarly, we need to calculate the moles of KOH present in the solution by using the pH of KOH solution.

pH = 14 - pOH

12 = 14 - pOH

pOH = 2

[OH-] = 10^-2 M

Number of moles of KOH = Molarity x Volume(in L) = 10^-2 x 3/1000 = 3 x 10^-5 moles

Step 3: Determine the limiting reagent

We need to determine the limiting reagent in the given mixture by comparing the moles of HCl and KOH obtained in the previous steps.

The limiting reagent is the one that is completely consumed in the reaction and decides the amount of product formed. The reagent that is present in lesser amount is the limiting reagent.

Here, HCl is the limiting reagent as it is present in lesser amount.

Step 4: Calculate the moles of water formed

The balanced chemical equation for the reaction between HCl and KOH is:

HCl + KOH → KCl + H2O

1 mole of HCl reacts with 1 mole of KOH to produce 1 mole of water.

Therefore, number of moles of water formed = 2 x 10^-5 moles (as HCl is the limiting reagent)

Step 5: Calculate the concentration of OH- ions in the solution

Number of moles of KOH left unreacted = 3 x 10^-5 - 2 x 10^-5 = 1 x 10^-5 moles

Concentration of KOH left unreacted = 1 x 10^-5 moles / 3 ml = 3.33 x 10^-3 M

Concentration of OH- ions formed from KOH = Concentration of KOH left unreacted = 3.33 x 10^-3 M

Step 6: Calculate the concentration of H+ ions in the solution

Number

Answer

Answer

The answer is 11.3

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