Question

Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.70.

You have in front of you

• 100 mL of 6.00×10⁻²M HCl,
• 100 mL of 5.00×10⁻²M NaOH, and
• plenty of distilled water.

You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 85.0 mL of HCl and 90.0 mL of NaOH left in their original containers.

ASSUMING THE FINAL SOLUTION WILL BE DILUTED TO 1.00L, HOW MUCH MORE HCl SHOULD YOU ADD TO ACHIEVE THE DESIRED PH?

Answer 1

Desired pH = 2.70

[H+] =10^–2.70 = 0.001995 M

Desired volume of HCl to be made = 1.0 L

Number of moles of H+ ions to be present in 1.0 L of solution

= M*V = 0.001995 M* 1.0 L = 0.001995 mol

Volume of HCl added first = 100 – 85 = 15 mL = 0.015 L

Volume of NaOH added = 100 – 90 = 10 mL = 0.010 L

Number of moles of HCl = M*V = 6.00×10⁻²M * 0.015 L = 0.0009 mol

Number of moles of NaOH = M*V = 5.00×10⁻²M * 0.010 L = 0.0005 mol

Net moles of HCl present in the resulting solution ( 15+10 = 25 mL) = 0.0009 – 0.0005 = 0.0004 mol

Number of moles of HCl need to be added = 0.001995 – 0.0004 = 0.001595 mol

Concentration of HCl stock solution = 6.00×10⁻²M

Volume of stock HCl solution needed for the remaining moles = moles/volume

= 0.001595 mol/ 6.00×10⁻²M = 0.026588 L = 26.58771 mL = 26.6 mL

Volume of HCl to be added = 26.6 mL

Volume of the solution that is mistakenly added NaOH (HCl + NaOH) = 25 mL

Volume of water be added to make 1.0 L of pH 2.70 = 1000 mL – (26.6 + 25) = 948.4 mL

Answer: Volume of HCl required = 26.6 mL