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pH problem

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1. 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 Thank you so much!
2. What have you tried?
3. (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]
4. (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)
5. (Original post by alow)
What have you tried?
I have attached pictures of how I attempted to solve the problem because the formatting goes a bit weird when I just type it out.
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