How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that has a ph of 8.53? kb for ammonia is 1.8*10-5. express your answer with the appropriate units?

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,

= 2.50 L * 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = - logKb,

= - log (1.8 * 10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log ([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73 = 5.37

[NH₄⁺ = 5.37 * 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,

mass of NH₄Cl = 10.74 mol * 53.5 g/mol

= 574.59 grams.