You have 3 beakers: Beaker 1 contains 50.0 ml of 0.123 M HC2H3O2 Beaker 2 contai
ID: 698878 • Letter: Y
Question
You have 3 beakers: Beaker 1 contains 50.0 ml of 0.123 M HC2H3O2 Beaker 2 contains 25.0 ml of 0.456 M NaC2H3O2 Beaker 3 contains 5.0 ml of 0.789 M HCl A.) What is the pH of the solution in beaker 1? B.) The contents of beaker 1 are poured into Beaker 2 and mixed. What is the pH of the resulting solution? C.) The contents of beaker 3 are now poured into the resulting solution. What is the new pH? D.) Could this final solution (part c) be an effective buffer? You have 3 beakers: Beaker 1 contains 50.0 ml of 0.123 M HC2H3O2 Beaker 2 contains 25.0 ml of 0.456 M NaC2H3O2 Beaker 3 contains 5.0 ml of 0.789 M HCl A.) What is the pH of the solution in beaker 1? B.) The contents of beaker 1 are poured into Beaker 2 and mixed. What is the pH of the resulting solution? C.) The contents of beaker 3 are now poured into the resulting solution. What is the new pH? D.) Could this final solution (part c) be an effective buffer? You have 3 beakers: Beaker 1 contains 50.0 ml of 0.123 M HC2H3O2 Beaker 2 contains 25.0 ml of 0.456 M NaC2H3O2 Beaker 3 contains 5.0 ml of 0.789 M HCl A.) What is the pH of the solution in beaker 1? B.) The contents of beaker 1 are poured into Beaker 2 and mixed. What is the pH of the resulting solution? C.) The contents of beaker 3 are now poured into the resulting solution. What is the new pH? D.) Could this final solution (part c) be an effective buffer?Explanation / Answer
A.
CH3COOH <-------> H+ + CH3COO-
IC: 0.123 0 0
C: - x + x + x
EC: 0.123 – x x x
Ka = [H+] [CH3COO-] / [CH3COOH]
Ka of CH3COOH = 1.8 x 10-5
So,
1.8 x 10-5 = [H+] [CH3COO-] / [CH3COOH]
1.8 x 10-5 = (x) (x) / (0.123 –x)
1.8 x 10-5 = x2 / (0.123) [since Ka is very small, x term in denominator can be discarded)
x2 = 2.2 x 10-6
x = 1.48 x 10-3
So, [H+] = x = 1.48 x 10-3
pH = - log[H+]
= - log (1.48 x 10-3)
= 2.83
(B)
50.0 mL of 0.123 M CH3COOH
Moles of CH3COOH = 0.123 M x 0.050 L = 0.00615 moles
25.0 mL of 0.456 M CH3COONa
Moles of CH3COONa = 0.456 M x 0.025 L = 0.0114 moles
Total volume = 50 mL + 25 mL = 75 mL = 0.075 L
[CH3COOH] = 0.00615 moles / 0.075 L = 0.082 M
[CH3COONa] = 0.0114 moles / 0.075 L = 0.152 M
Using Henderson-Hessalbalach equation
pH = pKa + log {[CH3COONa] / [CH3COOH] }
Ka of CH3COOH = 1.8 x 10-5
pKa = - log Ka
= - log (1.8 x 10-5)
= 4.74
pH = pKa + log {[CH3COONa] / [CH3COOH] }
pH = 4.74 + log {0.152 / 0.082 }
pH = 4.74 + log (1.85)
pH = 4.74 + 0.27
pH = 5.01
(c)
[CH3COOH] = 0.082 M
[CH3COONa] = 0.152 M
Volume = 75 mL
So,
Moles of CH3COOH = 0.082 M x 0.075 L = 0.00615 mole
Moles of CH3COONa = 0.152 M x 0.075 L = 0.0114 moles
5.0 mL of 0.789 M HCl
Moles of HCl = 0.789 M x 0.005 L = 0.003945 moles
So, 0.003945 moles of HCl will react with 0.003945 moles of CH3COONa to produce 0.003945 moles of CH3COOH. So, moles of CH3COONa will decrease and moles of CH3COOH will increase.
Moles after addition of HCl
Moles of CH3COOH = 0.00615 mole + 0.003945 moles = 0.010095 moles
Moles of CH3COONa = 0.0114 moles - 0.003945 moles = 0.007455 moles
Total volume = 75 mL + 5 mL = 80 mL = 0.080 L
So,
[CH3COOH] = 0.010095 moles / 0.080 L = 0.1261875 M
[CH3COONa] = 0.007455 moles / 0.080 L = 0.0931875 M
Using Henderson-Hessalbalach equation
pH = pKa + log {[CH3COONa] / [CH3COOH] }
Ka of CH3COOH = 1.8 x 10-5
pKa = - log Ka
= - log (1.8 x 10-5)
= 4.74
pH = pKa + log {[CH3COONa] / [CH3COOH] }
pH = 4.74 + log {0.0931875 / 0.1261875 }
pH = 4.74 + log (0.738484398)
pH = 4.74 - 0.13
pH = 4.61
(d)
Yes, this final solution is still buffer. Since pH < pKa.
Hence, it has not reached the half equivalent point.
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