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    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!
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    What have you tried?
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    (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]
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    (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)
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    (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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