Question

Imagine that you are in chemistry lab and need to make 1.00L of a solution with a pH of 2.40.

You have in front of you

100 mL of 6.00

Answer 1

M1 = 6*10^-2 HCl

M2 = 5*10^-2 NaOH

VT = 1 L

Since we have a the end:

84 ml of HCl

85 ml of NAOH

calculate the amounts used

100-84 = 16 ml of HCl used

100-85 = 15 ml of NaOH used

Calculate how many moles were there

M1V1 = 0.06M * 16 ml = 0.96 mmol of HCl

M2V2 = 0.05M * 15 ml = 0.75 mmol of NaOH

Therefore,

1 mol of HCL neutralize 1 mol of MaOH

0.96-0.75 = 0.21 mmol of HCl left in the V1+V2 solution = 16+15 = 31 ml

0.21 mmol of HCl

If we want pH = 2.4

That means

pH = -log[H+]

2.4 = -log[H+]

[H+] = 10^-2.4 = 0.00398

That is in 1 L

therefore, 0.00398 moles of HCl are needed

M1*(V?) = 0.00398 moles

NOte that we have already 0.21 mmol = 0.00021 mol of H+

0.00398 mol - 0.00021 mol = 0.00377 mol are needed

Since we have acid:

M1*V1* = 0.06M * V1 = 0.00377

V1 = 0.00377/0.06 = 62.83 ml

Add another 62.83 ml of Acid

31 ml + 62.83 ml + x ml of water = 1L

ml of Water = 906.17 ml

ml of Acid in TOTAL = 16+62.83 = 78.83 ml were used in total

But the correct answe is, how much after the addition of NaOH an dHcl

that is:

62.83 ml of Acid