1.A buffer solution contains 0.434 M nitrous acid and 0.239 M potassium nitrite
ID: 1010179 • Letter: 1
Question
1.A buffer solution contains 0.434 M nitrous acid and 0.239 M potassium nitrite . If 0.0333 moles of sodium hydroxide are added to 150 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding sodium hydroxide. ) pH=
2.A buffer solution contains 0.394 M ammonium chloride and 0.492 M ammonia.
If 0.0179 moles of hydrochloric acid are added to 150 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume change does not change upon adding hydrochloric acid) pH =
3.How many moles of sodium hydroxide would have to be added to 125 mL of a 0.415 M acetic acid solution, in order to prepare a buffer with a pH of 4.660?
4.An aqueous solution contains 0.397 M hydrocyanic acid. How many mL of 0.383 M sodium hydroxide would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 9.180?
5.How many grams of solid ammonium bromide should be added to 0.500 L of a 0.266 M ammonia solution to prepare a buffer with a pH of 9.880 ? grams ammonium bromide =
6.How many grams of solid sodium fluoride should be added to 2.00 L of a 0.168 M hydrofluoric acid solution to prepare a buffer with a pH of 2.472 ? grams sodium fluoride =
Explanation / Answer
.
1) Pka of HNO2/KNO2 =3.39
Moles of HNO2 in the buffer=0.434 mol/L*0.150L=0.065 moles
Moles of NO2-=0.239mol/L*0.15L=0.036 moles
If 0.0333 moles of NaOH is added it will neutralise 0.0333 moles of acid ,HNO2,remaining HNO2=0.065-0.0333=0.0317 moles
Moles of NO2- will increase as 0.0333 moles NaNO2 will be formed.=0.0333+0.036=0.0693 moles
pH=pka+log [base]/[acid]{henderson -hasselbach equation}
=3.39+log (0.0693/0.0317)=3.39+0.34=3.73
pH=3.73
2.
Moles of amm.chloride=0.394 mol/L*0.15L=0.059 moles
Moles of ammonia=0.492 mol/L*0.15=0.074 moles
After adding HCl,moles of base (ammonia neutralized)=0.0179 moles
Moles of ammonia remaining=0.074-0.0179=0.056 moles
Moles of amm.chloride=0.059+0.0179=0.077 moles
pOH=pkb + log [acid]/[base]=4.75+log(0.077/0.056)=4.75
pH=14-pOH=14-4.75=9.25
3.
Moles of acetic acid=0.415mol/L*0.125 L=0.0519 moles
pH=pka+log [base]/[acid]
Pka acetic acid/acetate=4.75
4.660=4.75+log [base]/[acid]
-0.09=log[base]/[acid]
[base]/[acid]=0.91
[acid]+[base]=0.0519 moles
[acid]=(1/1.91)*0.0519=0.0271 moles
[base]=(0.91/1.91)*0.0519=0.0247 moles
So amount of acid needs to be neutralized=0.0247 moles=moles of NaOH added
4) similar to 3)
Pka of HF/F-=3.14
pH=pka+ log [F-]/[HF]
2.472=3.14+log [F-]/[HF]
-0.067=log [F-]/[HF]
[F-]/[HF]=0.215
[HF]=0.168 mol/L*2.00L=0.336 moles
[F-]=(0.215)*0.336=0.0722 moles
So 0.0722 moles of F- needs to be added=0.0722 moles*its molar mass=0.0722 moles*41.99 g/mol=3.033 grams
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