Please help and show work!! B.)What is the pH of a buffer system prepared by dis
ID: 965337 • Letter: P
Question
Please help and show work!!
B.)What is the pH of a buffer system prepared by dissolving 10.70 grams of NH4Cl and 40.00 mL of 12 M NH3 in enough water to make 1.000 L of solution? K b = 1.80 × 10-5 for NH3.
What is the pH of a buffer system prepared by dissolving 10.70 grams of NH4Cl and 40.00 mL of 12 M NH3 in enough water to make 1.000 L of solution? K b = 1.80 × 10-5 for NH3.
C.)What is the pH of a solution prepared by mixing 100.00 mL of 0.020 M Ca(OH)2 with 50.00 mL of 0.200 M NaOH? Assume that the volumes are additive.
What is the pH of a solution prepared by mixing 100.00 mL of 0.020 M Ca(OH)2 with 50.00 mL of 0.200 M NaOH? Assume that the volumes are additive.
5.22 8.78 9.73 4.27Explanation / Answer
a) Let's calculate the moles of each:
moles A = 0.040 * 0.1 = 0.004 moles
moles B = 0.1 * 0.01 = 0.001 moles
the moles of base are way smaller than acid, so the remaining moles after the reaction:
CH3COOH + OH- <---------> CH3COO- + H2O
0.004 0.001 0 0
0.003 0 0.001
With this, the pH can be calculated with the HH equation:
pH = pKa + logA/HA ---> pKa = -log(1.8x10-5)= 4.744
pH = 4.744 + log(0.001/0.003)
pH = 4.27
b) calculate first the Ka = 1x10-14 / 1.8x10-5 = 5.56x10-10
The moles of NH4Cl = 10.70 / (14+4+35.5) = 0.2 moles
moles of NH3 = 0.04 * 12 = 0.48 moles
NH4+ ---------> NH3 + H+
pH = -log(5.56x10-10) + log(0.48/0.2)
pH = 9.63
c) Both are strong bases so:
Ca(OH)2 ------> Ca2+ + 2OH-
[OH-] = 2*0.02 = 0.04 M ----> moles OH = 0.04 * 0.1 = 0.004 moles
NaOH ------> Na+ + OH-
[OH] = 0.2 M ---> moles = 0.2 * 0.05 = 0.01 moles
Total moles OH = 0.004+0.01 = 0.014 moles
[OH] = 0.014 / 0.15 = 0.093 M
pOH = -log(0.093) = 1.03
pH = 14-1.03 = 12.97
Try to do yourself part d) with the examples here. Hope this helps