How do i calculate the pH when 200ml of .200M HCl is added .4998M of sodium acetate?
This problem reads suspiciously like something is missing. ......of 0.200 M HCl is added something here .4998 M socium acetate. I don't know how much sodium acetate you have.
sorry huge typo
How do i calculate the pH when 200ml of .200M HCl is added to 200.0 ml of .4998M of sodium acetate?
THANK YOU. I've been looking at this thing, of and on, for over an hour. Even tried some dry runs to see it might work out if I simply assumed a volume of sodium acetate. Of course, none of them would pan out because it made a difference in how much I assumed. SO,
let's call sodium acetate NaAc and the acetate ion Ac^-. Helps on typing.
Ac^- + H^+ ==> HAc (acetic acid)
initial:
Ac^- = M x mL = 0.4998 x 200 mL = 99.96 millimoles.
H^+ = M x mL = 0.2 x 200 mL = 40 mmoles.
HAc = 0 mmoles.
final:
HAc = 40 mmoles.
H^+ = 0
Ac^- = 99.96-40 = 59.96
(Ac^-) = mmols/mL (or moles/L)
(HAc) = mmols/mL (or moles/L)
Substitute into HH equation. Base is Ac^- and acid is HAc.
I get something like 4.9 or so but that's just a close estimate. You need to do it more accurately. By the way, the mL in the concn conversion is 400 mL.
To calculate the pH of a solution when mixing a strong acid (HCl) with a weak base (sodium acetate), you need to determine the concentrations of the relevant ions in the final solution.
Step 1: Write the balanced chemical equation for the reaction between HCl and sodium acetate:
HCl + CH3COONa <--> CH3COOH + NaCl
Step 2: Calculate the moles of HCl added:
Moles of HCl = Volume of HCl (in L) x Concentration of HCl (in mol/L)
Moles of HCl = 0.200 L x 0.200 mol/L = 0.040 mol
Step 3: Calculate the moles of sodium acetate:
Moles of sodium acetate = Volume of sodium acetate (in L) x Concentration of sodium acetate (in mol/L)
Moles of sodium acetate = 0.200 L x 0.4998 mol/L = 0.09996 mol
Step 4: Determine the reaction that occurs between HCl and sodium acetate:
Since HCl is a strong acid, it will fully dissociate into H+ and Cl- ions. However, sodium acetate is a weak base, so it will only partially dissociate into Na+ and CH3COO- ions.
The reaction will consume H+ ions and produce CH3COOH. So, the moles of H+ ions consumed will be equal to the moles of sodium acetate added.
Step 5: Calculate the remaining concentration of H+ ions:
Remaining moles of H+ ions = Moles of HCl - Moles of H+ ions consumed
Remaining moles of H+ ions = 0.040 mol - 0.09996 mol = -0.05996 mol (negative because it was consumed)
Step 6: Calculate the molar concentration of H+ ions:
Molar concentration of H+ ions = Remaining moles of H+ ions / Total volume of the solution (in L)
Molar concentration of H+ ions = -0.05996 mol / 0.200 L = -0.2998 mol/L (negative because it was consumed)
Step 7: Calculate the pOH:
pOH = -log10[OH-] (Since sodium acetate is a weak base, we need to find OH- concentration)
OH- concentration = Remaining moles of sodium acetate / Total volume of the solution (in L)
OH- concentration = 0.09996 mol / 0.200 L = 0.4998 mol/L
pOH = -log10(0.4998) ≈ 0.301
Step 8: Calculate the pH:
pH = 14 - pOH
pH = 14 - 0.301 ≈ 13.699
Therefore, the pH of the solution is approximately 13.699.