1. How many grams of dry NH4Cl need to be added to 1.50 L of a 0.200 M solution

Question

1. How many grams of dry NH4Cl need to be added to 1.50 L of a 0.200 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.60? Kb for ammonia is 1.8e-5

2 Part A:.What is the pH of a buffer prepared by adding 0.45 mol of the waek acid HA to 0.608 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66e-7

3.Part B: What is the pH after 0.150mol of HCl is added to the buffer from part A? Assume no volume change on the addition of the acid.

4.Part C: What is the pH after 0.195mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

Explanation / Answer

1) pH = pKa + log(base/acid)

8.60 = 9.255 + log([base]/[acid])

moles of base (ammonia) = (0.2M)*(1.5L) = 0.3 moles

moles of acid (NH4Cl) = x

When x moles of acid are added, x moles of base are converted to acid

8.60 = 9.255 + log((0.3 - x)/x)

-0.655 = log((0.3 - x)/x)

10^-0.655 = (0.3 - x)/x

Solve for x:

x = 0.24564 moles

(0.25464 moles)(53.491 grams/mole) = 13.14 grams NH4Cl

2) pH = pKa + log(base/acid) = 6.247 + log((0.608 - 0.45)/0.45)

pH = 5.79

3) pH = 6.247 + log((0.608 - 0.45 - 0.15)/(0.45 + 0.15))

pH = 6.247 - log(0.008/0.6)

pH = 4.37

4) pH = 6.247 + log((0.008 + 0.195)/(0.6 - 0.195))

pH = 5.95

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