Question

Can somebody please help with this, I do not know what else to do. How many grams of dry ice NH4Cl need to be added to 2.50 L of a 0.300 mol solution of ammonia NH3, to prepare a buffer solution that has a pH of 8.85? Kb for ammonia is 1.8*10^-5

Answer 1

first you need to know the molarity of \(NH_4Cl\), which is the same as saying \([NH_4^+]\).

so you use the henderson-hasselbalch equation: \(pH=pKa+log\dfrac{[A]}{[AH^+]}\); you know: \(pK_b=-log(K_b)\) and \(pK_a+pK_b=14\)

so, \(pK_a=14-pK_b=14-(-log(K_b))=14+log(K_b)\)

so you have: \(pH=14+log(K_b)+log\dfrac{[A]}{[AH^+]}\)

\([A]=[NH_3]\) and \([AH^+]=[NH_4^+]\)

\(pH=14+log(K_b)+log\dfrac{[NH_3]}{\color{red}{[NH_4^+]}}\)

Answer 2

once you find the molarity, convert to moles, then to grams.