Question

Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.50.

You have in front of you

• 100 mL of 7.00×10⁻² M HCl,
• 100 mL of 5.00×10⁻² M 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 80.0 mL of HCl and 90.0 mL of NaOH left in their original containers.

Assuming the final solution will be diluted to 1.00 L , how much more HCl should you add to achieve the desired pH?

Express your answer to three significant figures and include the appropriate units.

Answer 1

Volume of HCl added = Intial volume of HCl - volume of HCl left

= 100 mL - 80 mL

20 mL

Volume of NaOH added = Intial volume of NaOH - volume of NaOH left

= 100 mL - 90 mL

= 10 mL

Required pH of final solution = 2.5

pH = -log [H⁺]

-log [H⁺] = 2.5

[H⁺] =10^(-2.5) = 0.00317 M = 0.00317 mol/L

Therefore, we need a solution with 0.00317 moles of H⁺ ions in 1 L.

Volume of HCl added = 20 mL

Concentration of HCl solution = 7*10^-2 M = 0.07 M= 0.07 mol / L

No. of Moles of HCl added =( 0.07 moles / 1000 mL ) * 20 mL ----------[1L =1000mL]

=0.0014 moles

Volume of NaOH added = 10 mL

Concentration of NaOH solution = 5*10^-2 M = 0.05 M= 0.05 mol / L

No. of Moles of NaOH  added =( 0.05 moles / 1000 mL ) * 10 mL ----------[1L =1000mL]

=0.0005 moles

The balanced equation for neutralization reaction between HCl and NaOH is

HCl (aq) +  NaOH(aq) $\rightarrow$ NaCl(aq) + H₂O(l)

Therefore, 1 mole of NaOH neutralizes 1 mole of HCl.

0.0005 moles of NaOH will neutralize 0.0005 moles of HCl.

No. of moles of HCl left after neutralization = 0.0014 moles - 0.0005 moles

=0.0009 moles

No. of moles of HCl required to be added to obtain the desired pH =

No. of moles of H⁺ required - no. of moles of HCl left after neutralizaion in solution

= 0.00317 mol/L - 0.0009 moles

=0.00226 moles

Concentration of HCl solution = 0.07 M = 0.07 mol/L

Therefore, 0.07 moles of HCl are present in 1L(1000 mL ) solution.

Volume of 0.07 M HCl containing 0.00226 moles HCl = (1000 mL/ 0.07 moles ) * 0.00226 moles

= 32.29 mL

= 32.3 mL

Volume of HCl required to be added = 32.3 mL