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Ph calculations IAL

Adhiiiiii

Calculate the pH of a mixture of 20.0 cm³ of 1.00 mol dm³ HCl(aq) and 5 cm³ of 1.00 mol dm³ NaOH(aq)

Reply 1

Pigster

Original post by Adhiiiiii

Calculate the pH of a mixture of 20.0 cm³ of 1.00 mol dm³ HCl(aq) and 5 cm³ of 1.00 mol dm³ NaOH(aq)

What have you tried, so far?

Reply 2

Nitrotoluene

Dear Adhiiiiii
I find it puzzling that you do not share your efforts in solving chemical problems.
This way you will never develop a deeper understanding of chemistry and problem solving skills.
Unfortunately, relying on others to solve problems can hinder your progress and prevent you from grasping the basic concepts.

To calculate the pH of the mixture, we must first calculate the number of moles of HCl and NaOH and then calculate the number of moles of H+ ions in the mixture.
1: Calculate the number of moles of HCl
Molarity of HCl = 1.00 mol/dm^3, volume of HCl = 20.0 cm^3 = 0.0200 dm^3 (conversion from cm^3 to dm^3)
Number of moles of HCl = molar mass x volume = 1.00 mol/dm^3 x 0.0200 dm^3 = 0.0200 mol

2: Calculate the number of moles of NaOH
Molarity of NaOH = 1.00 mol/dm^3, volume of NaOH = 5 cm^3 = 0.0050 dm^3 (conversion from cm^3 to dm^3)
Number of moles of NaOH = molarity x volume = 1.00 mol/dm^3 x 0.0050 dm^3 = 0.0050 mol

3: Calculate the number of moles of H+ ions in the mixture.
The reaction between HCl and NaOH is as follows

HCl + NaOH = NaCl + H2O

From this reaction we can see that 1 mole of HCl reacts with 1 mole of NaOH to form 1 mole of H+ ions.
Since we have 0.0200 moles of HCl and 0.0050 moles of NaOH, the number of moles of H+ ions in the mixture is equal to :
Number of moles of H+ ions = Number of moles of HCl - Number of moles of NaOH = 0.0200 mol - 0.0050 mol = 0.0150 mol

4: Calculate the concentration of H+ ions in the mixture
Volume of the mixture = volume of HCl + volume of NaOH = 20.0 cm^3 + 5 cm^3 = 25.0 cm^3 = 0.0250 dm^3 (conversion from cm^3 to dm^3)
H+ ion concentration = number of moles of H+ ions / volume of the mixture = 0.0150 mol / 0.0250 dm^3 = 0.600 mol/dm^3

5: Calculate the pH of the mixture
pH = -log base 10[H+] = -log base 10(0.600 mol/dm^3) = 0.22

The pH of the mixture is therefore 0.22.