0.487 M KBrO 35.00 mL 0.264 M HNO3 Kb 4.0×10^-6

0.487 M KBrO 35.00 mL 0.264 M HNO3 Kb 4.0×10^-6

 

 

A 35.00 mL sample of 0.487 M KBrO

is titrated with 0.264 M HNO3. (Kb BrO^- = 4.0×10^-6)

a) Write a balanced net ionic equation for the reaction.

b) How many milliliters of HCl are required to reach the equivalence point?

c) What is the pH at the equivalence point?

d) Calculate [K^+], [NO3^-], [H^+], [BrO^-], and [HBrO]

at the equivalence point. (Assume volumes are additive.)

 

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▶ 참고: 약염기-강산 적정 [ https://ywpop.tistory.com/2742 ]

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a) Write a balanced net ionic equation for the reaction.

KBrO + HNO3 → HBrO + KNO3

 

 

 

 

b) How many milliliters of HCl are required to reach the equivalence point?

MV = M’V’

( 참고 https://ywpop.tistory.com/4689 )

 

(0.487 M) (35.00 mL) = (0.264 M) (? mL)

? = (0.487) (35.00) / (0.264) = 64.56 mL

 

 

 

 

c) What is the pH at the equivalence point?

HBrO(aq) ⇌ H^+(aq) + BrO^-(aq)

 

 

 

Ka = Kw / Kb

= (10^(-14)) / (4.0×10^(-6))

= 2.5×10^(-9)

 

 

 

생성된 HBrO의 몰수 = 처음 약염기의 몰수 = 가한 강산의 몰수

(0.487 mol/L) (35.00/1000 L) = 0.017045 mol HBrO

 

0.017045 mol / [(35.00+64.56)/1000 L]

= 0.1712 M HBrO

( 참고 https://ywpop.tistory.com/2742 )

 

 

 

x = [Ka × C]^(1/2)

( 참고 https://ywpop.tistory.com/4294 )

 

= [ (2.5×10^(-9)) × 0.1712 ]^(1/2)

= 2.0688×10^(-5) M = [H^+]

 

 

 

pH = -log[H^+]

= -log(2.0688×10^(-5))

= 4.68

 

 

 

 

d) Calculate [K^+], [NO3^-], [H^+], [BrO^-], and [HBrO]

at the equivalence point. (Assume volumes are additive.)

 

[K^+] = 0.487 M (구경꾼 이온)

 

[NO3^-] = 0.264 M (구경꾼 이온)

 

[H^+] = 2.07×10^(-5) M = [BrO^-]

 

[HBrO] = 0.171 M

 

 

 

 

[키워드] KBrO + HNO3 적정 기준문서, 약염기 + 강산 적정 기준문서