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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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