A weak acid (HA) can be used to make a buffer for biochemical experiments: HA=H++Aa. Calculate the pH of a solution made by adding 0.060 moles of HA to 250mLs of de-ionized water. 4 2 = 0.01mal 0.250 ** ** o 0.24M HA=H" +A" ka. 10 phe = 10 46 = 5.47 × 105 3.47 × 10°5 = X -x 5.47 × 105 (024-x) = x² 8.33×108 0.24 b. Now suppose 0.020 moles of strong acid (NaOH) is added to the solution from part a. Calculate the new pH (you may -4 ignore dilution). moles of 4 = 2.9 × 10 *0.25 = 7.25 × 10* M HA + NaOH - NaA+H2O moles

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Chapter15: Acid-base Equilibria
Section: Chapter Questions
Problem 8ALQ: You have a solution of the weak acid HA and add some of the salt NaA to it. What are the major...
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A weak acid (HA) can be used to make a buffer for biochemical experiments: HA=H++Aa. Calculate the pH of a solution
made by adding 0.060 moles of HA to 250mLs of de-ionized water.
4
2
=
0.01mal
0.250
o 0.24M HA = H+A ka. 10 phe
= 10
46 = 5.47 × 105 3.47×105
x
=
-x 5.47 × 10³ (024-x)=x8.33×10
5.47×10(024-x)
2
-8
-
0.24
b. Now suppose 0.020 moles of strong acid (NaOH) is added to the solution from part a. Calculate the new pH (you may
ignore dilution). moles of 4 = 2.9 × 103 *0.25 = 7.25×10^ M HA + NaOH
→
NaA+H₂O moles
2
[[HA] = 0.06mol
= 6.04mol
{
0.04(mol)
[HA] =
= 0.16M
0.252
-0.02mol ]
-3
257 x 10
-4
moles of N
= 7.175 × 10 7.175×10 +0.02 → = 0.020715mol
4
0.020725
[A]= = 0.0829m PH = p, a + log(B) = 4.46 + log(0.082) = 4.17 Please let me know if my work is correct. If not,
0.25(L)
please demonstrate the proper way.
A
0.16
Transcribed Image Text:A weak acid (HA) can be used to make a buffer for biochemical experiments: HA=H++Aa. Calculate the pH of a solution made by adding 0.060 moles of HA to 250mLs of de-ionized water. 4 2 = 0.01mal 0.250 o 0.24M HA = H+A ka. 10 phe = 10 46 = 5.47 × 105 3.47×105 x = -x 5.47 × 10³ (024-x)=x8.33×10 5.47×10(024-x) 2 -8 - 0.24 b. Now suppose 0.020 moles of strong acid (NaOH) is added to the solution from part a. Calculate the new pH (you may ignore dilution). moles of 4 = 2.9 × 103 *0.25 = 7.25×10^ M HA + NaOH → NaA+H₂O moles 2 [[HA] = 0.06mol = 6.04mol { 0.04(mol) [HA] = = 0.16M 0.252 -0.02mol ] -3 257 x 10 -4 moles of N = 7.175 × 10 7.175×10 +0.02 → = 0.020715mol 4 0.020725 [A]= = 0.0829m PH = p, a + log(B) = 4.46 + log(0.082) = 4.17 Please let me know if my work is correct. If not, 0.25(L) please demonstrate the proper way. A 0.16
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