A 20.0 mL sample of 0.20 M sodium acetate is titrated with 0.11 M HCl)aq). What is pH after the addition of50.0 mL HCl(aq)? Kbof CH3CO2^- = 5.6x10^-10)
So far I have the
Initial amount of sodium acetate (0.02)(0.20) = .004 mol
Amount of HCl added (0.05)(0.11) = 0.0055 mol
Amount of sodium aceatate after reaction(0.04) - (0.0055) = -.0015
So there is less than nothing left. That can't be right
You are ok so far except that you wrote (0.04)-(0.0055) = -.0015 when you should have written
(0.004)NaAc - (0.0055)HCl = but I think that was just a typo. I have typed in the NaAc and HCl part to clarify things. So what has happened is that you have no NaAc left (technically no Ac^- left except what little may be present from the ionization of a weak acid like acetic acid--which is decreased even further by the addition of the strong acid, HCl). So in effect you have 0.0015 moles HCl in 70 mL water (0.070 L) which means the molarity of HCl is ?? and pH is ??.
Molarity = mol/L
0.0015/.070 = 0.0214
-log 2.14x10^-2 = 1.67
It seems like there is a mistake in your calculation. Let's go through the problem again step by step to find the correct answer.
1. Calculate the initial amount of sodium acetate:
Volume of sodium acetate = 20.0 mL = 0.020 L (remember to convert mL to L)
Concentration of sodium acetate = 0.20 M
Amount of sodium acetate = Volume x Concentration = 0.020 L x 0.20 M = 0.004 mol
Your calculation for the initial amount of sodium acetate is correct.
2. Calculate the amount of HCl added:
Volume of HCl = 50.0 mL = 0.050 L (convert mL to L)
Concentration of HCl = 0.11 M
Amount of HCl = Volume x Concentration = 0.050 L x 0.11 M = 0.0055 mol
Your calculation for the amount of HCl added is also correct.
3. Calculate the amount of sodium acetate remaining after the reaction:
Initial amount of sodium acetate - Amount of HCl added = 0.004 mol - 0.0055 mol = -0.0015 mol
It seems like the result is negative, which implies that there is less sodium acetate remaining than there was initially. However, negative values do not make sense in this context. This discrepancy may arise from the assumption that the reaction between sodium acetate and HCl goes to completion, but in reality, it does not.
To find the pH of the solution, we need to consider the dissociation of sodium acetate (CH3CO2-) and the reaction with HCl to form acetic acid (CH3COOH) and NaCl.
The dissociation of sodium acetate can be represented by the following equation:
CH3CO2- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq)
Since the concentration of sodium acetate is initially 0.20 M, and we can assume complete ionization, the concentration of CH3CO2- can be considered as 0.20 M.
4. Calculate the concentration of OH- ions produced by the dissociation of sodium acetate:
CH3CO2- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq)
At equilibrium, the concentration of OH- ions can be calculated using the equilibrium constant (Kb) for the reaction:
Kb = [CH3COOH][OH-] / [CH3CO2-]
Given that Kb for CH3CO2- is 5.6 x 10^-10, and the concentration of CH3CO2- is 0.20 M, we can rearrange the equation to solve for [OH-]:
[OH-] = (Kb x [CH3CO2-]) / [CH3COOH]
[OH-] = (5.6 x 10^-10) x (0.20) / [CH3COOH]
Now, we need to find the concentration of acetic acid (CH3COOH). Since HCl reacts with sodium acetate to form acetic acid, we need to calculate the moles of acetic acid formed.
5. Calculate the moles of acetic acid formed through the reaction with HCl:
The balanced chemical equation for the reaction is:
CH3CO2- (aq) + HCl (aq) ⇌ CH3COOH (aq) + Cl- (aq)
From the equation, we can see that one mole of CH3CO2- reacts with one mole of HCl to form one mole of CH3COOH. Since the initial amount of sodium acetate is 0.004 mol, the moles of acetic acid formed will also be 0.004 mol.
6. Calculate the concentration of acetic acid (CH3COOH):
Volume of acetic acid = (20.0 mL + 50.0 mL) = 70.0 mL = 0.070 L (convert mL to L)
Concentration of acetic acid = (moles of acetic acid) / (volume of acetic acid in L) = 0.004 mol / 0.070 L ≈ 0.057 M
Now that we have the concentration of acetic acid (CH3COOH), we can proceed to find the concentration of OH- ions and then the pOH of the solution.
7. Calculate the concentration of OH- ions:
[OH-] = (5.6 x 10^-10) x (0.20 M) / (0.057 M)
Calculating this expression will give you the concentration of OH- ions in the solution.
8. Calculate the pOH of the solution:
pOH = -log10[OH-]
Now that you have the concentration of OH- ions, you can use this value to calculate the pOH of the solution.
9. Calculate the pH of the solution:
pH + pOH = 14
Using the pOH value, you can calculate the pH of the solution.