1. You make a buffer solution containing 5.342 g of NaF and 3.245 g of HF (pKa=
ID: 553242 • Letter: 1
Question
1. You make a buffer solution containing 5.342 g of NaF and 3.245 g of HF (pKa= 3.17) in 1.00 L of H2O. What is the approximate pH of the buffer?
A) 0.50 B) 1.00 C) 3.06 D) 3.17 E) 3.28
2. Methylamine is a weak base with pKb= 3.36. Calculate the pH of a 0.010M solution of methylamine in water.
A) 3.36 B) 7.00 C) 10.64 D) 11.26 E) 12.05
3. The Li+ ion has a hydrated radius of 600 pm. What is the activity of the ion in a 0.350 M solution of LiCl?
A) 0.0100 B) 0.0256 C) 0.254 D) 0.73 E) 0.95
4. You add 0.0500 mol of acetic acid (pKa= 4.76) to a 250.0 mL volumetric ask. How many mL of 0.500 M NaOH should you add to make a bu er with pH =5.00?
A) 63.50 mL B) 25.60 mL C) 5.23 mL D) 1.85 mL E) 0.0635 mL
5. You have a 6.5X10^-7M solution of sodium formate (Na^+HCOO^-) in water. Formate ion is the conjugate base of formic acid (HCOOH). What is the charge balance for the solution?
A) [H3O+]=[HCOO-] B) 6.5x10^-7 M = [HCOOH] + [HCOO-] C) [Na+] + [H3O+]=[HCOO-] + [OH-] D) [Na+]=[OH] E) None of the above.
Explanation / Answer
1)
Molar mass of NaF = 1*MM(Na) + 1*MM(F)
= 1*22.99 + 1*19.0
= 41.99 g/mol
mass of NaF = 5.342 g
we have below equation to be used:
number of mol of NaF,
n = mass of NaF/molar mass of NaF
=(5.342 g)/(41.99 g/mol)
= 0.1272 mol
volume , V = 1 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 0.1272/1
= 0.1272 M
So,
[NaF] = 0.1272 M
Molar mass of HF = 1*MM(H) + 1*MM(F)
= 1*1.008 + 1*19.0
= 20.008 g/mol
mass of HF = 3.245 g
we have below equation to be used:
number of mol of HF,
n = mass of HF/molar mass of HF
=(3.245 g)/(20.008 g/mol)
= 0.1622 mol
volume , V = 1 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 0.1622/1
= 0.1622 M
[HF] = 0.1622 M
we have below equation to be used:
This is Henderson–Hasselbalch equation
pH = pKa + log {[conjugate base]/[acid]}
pH = pKa + log {[NaF]/[HF]}
= 3.17+ log {0.1272/0.1622}
= 3.064
Answer: 3.064
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