A student dissolved 1.75 g NH_4CI in enough 0.100 M aqueous NH_3 to make 250.0 m

Question

A student dissolved 1.75 g NH_4CI in enough 0.100 M aqueous NH_3 to make 250.0 mL of solution. a. Calculate the pH of this solution. b. When the student added a small amount of NaOH to this solution, she noticed that the pH did not change much. Explain why the addition of some NaOH did not significantly affect the pH and provide a balanced chemical equation to support your answer. c. When the student added a small amount of HCI to this solution, she noticed that the pH did not change much. Explain why the addition of some HCI did not significantly affect the pH and provide a balanced chemical equation to support your answer. When making a buffer solution we use a weak conjugate acid-base pair. Explain why the conjugate base of a strong acid would not work in a buffer solution. To support your answer, provide a balanced chemical equation showing why Cl^1- which is the conjugate base of the strong acid HCI, will not work as part of a buffer.

Explanation / Answer

Q1.

m = 1.75 g of NH4Cl

mol = mass/MW = 1.75/53.491 = 0.0327157 mol of NH4+

mol of NH3 = MV = 0.1*0.25= 0.025 mol

this is a buffer since NH3 and NH4+ (conjugate) is present

pKa = 14 - pKb = 14-4.75 = 9.25

pH= pKa + log(NH4+/NH3)

pH= 9.25+ log(0.0327157 /0.025 )

pH = 9.3668

b)

when we add OH-, there buffer pH will not hcange, since it is neutralized by

NH4+ + OH- --> NH3 + H3O

c)

when we add H+ , there is no chang ein pH since the NH3 neutralizes the reation

NH3 + H+ --> NH4+ forms

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