pH problem

How much water and vinegar (the vinegar is 4% acetic acid) do I need to create a 300 mL solution of pH 3? Please show your working out so I can understand the problem, and the sooner the better because this is due Monday :redface:

Thank you so much! :smile:

What have you tried?

OP

Original post by alow

What have you tried?

This is how I attempted to solve the problem (with help from a friend), but it's a bit complicated and so I thought it would be nice to check if this gave the right answer and just get a second opinion. By the way, when I originally did these calculations I thought I would need 1.5 L of water so the calculations are for 1.5 L, not 300 mL.pH 3: pH= 3pH = -log[H3O+][H3O+] = 10-3 M
CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+Ka=1.7610-5M
At equilibrium : y = CH3COOH molarityCH3COOH = y - 10-3 MCH3COOH = [H3O+]=10-3 M
Ka= [conjugate base] [H+]/[conjugate acid]
Ka= [CH3COO-] [H3O+]/[CH3COOH]

Reply 3

EmmaChristina09

OP

Original post by EmmaChristina09

This is how I attempted to solve the problem (with help from a friend), but it's a bit complicated and so I thought it would be nice to check if this gave the right answer and just get a second opinion. By the way, when I originally did these calculations I thought I would need 1.5 L of water so the calculations are for 1.5 L, not 300 mL.pH 3: pH= 3pH = -log[H3O+][H3O+] = 10-3 M
CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+Ka=1.7610-5M
At equilibrium : y = CH3COOH molarityCH3COOH = y - 10-3 MCH3COOH = [H3O+]=10-3 M
Ka= [conjugate base] [H+]/[conjugate acid]
Ka= [CH3COO-] [H3O+]/[CH3COOH]

(continued)
1.76 × 10⁻⁵= (10⁻3)² / (y - 10⁻3)y - 10⁻3 = (10⁻3)² / (1.76 × 10-5)y- 10⁻3 =0.0568y =0.0578
0.0578 M CH₃COOH is needed.
If I want a 2 L solution:M = n/vn=Mvn=(0.0578 mol/L) × (2 L)n=0.1156
Molar mass of CH₃COOH = (12.012) + (1.01 4) + (16.002) = 60.06 g/mol n =mmm(0.116) (60.06) = 6.95
I need 6.95 g of acetic acid.
Since the vinegar is 4% acetic acid, I will need:4100=6.95 gramsx grams of vinegarx = 173.6 grams of vinegar
D = m/v
Density of acetic acid = 1.045 1.045 = 6.95/ vv = 6.646 mL
4100=6.646 mLy mL of vinegary = 166.15Amount of water needed = 2,000 mL - 166.15 mL= 1833.85 mL of water
∴ I will need:173.6 grams of vinegar1833.85 mL of waterOr 28.39 (of 300 mL water)