Given four beakers labeled A, B, C, and D. Add 40 mL of distilled water to beake
ID: 960055 • Letter: G
Question
Given four beakers labeled A, B, C, and D.
Add 40 mL of distilled water to beakers A and C.
Add 40 mL of the prepared sodium acetate/acetic acid buffer solution to both beakers B and D.
Pipette 1.0 mL of 6 M HCl, hydrochloric acid solution into beakers A and B.
Pipette 1.0 mL of 6 M NaOH, sodium hydroxide solution into beakers C and D.
If the concentration of CH3CO2Na after dilution is 0.62M and the concentration of CH3CO2H after dilution is 0.51M
Calculated pH of beaker A and B after the addition of 6 M HCl.(Ka of CH3CO2H=1.32x10^-4)
Explanation / Answer
pH in beaker A
40 mL distilled water.
1 mL 6M HCl added to 40 mL water.
Resulting pH =
1) Calculate total H+ in 40 mL of water:
(1.00 x 10¯7 mol/L) ( 0.04 L) = 4 * 10^-9 mol
2) Calculate total H+ in HCl:
(6 mol/L) (0.001 L) = 0.006 mol
3) Add the results together:
4 * 10^-9 mol + 0.006 mol = 0.0060 mol
4) Calculate new molarity of hydrogen ion:
0.0060 mol / 41 L = 0.00014 M
5) Calculate new pH:
- log 0.00014 = 3.834
pH in beaker B
First write the equation for the ionization of acetic acid in water and the related Ka expression rearranged to solve for the hydronium ion concentration.
CH3COOH(aq) + H2O(l) à H3O+(aq) + CH3COO-(aq)
[H3O+] = Ka[CH3COOH] / [CH3COO-]
Second, we will make an "ICE" chart. Let "x" represent the hydronium ion concentration once equilibrium has been re-established. We will assume that all of the added acid is consumed.
CH3COOH H3O CH3COO-
I 0.51 (0.001L)(6M HCl)=0.006M 0.62
C +0.006 -0.006 -0.006
E 0.516 x 0.614
Substitute into the Ka expression and solve for the hydronium ion concentration. Convert the answer into pH.
[H3O+] = (1.32 * 10^-4)(0.516/0.614) = 0.00011
pH = 3.95
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