Question

5.

A. How many grams of solid potassium fluoride should be added to 2.00 L of a 0.212 M hydrofluoric acid solution to prepare a buffer with a pH of2.374 ?

grams potassium fluoride = _____g.

B. How many grams of solid sodium nitrite should be added to 0.500 L of a0.154 M nitrous acid solution to prepare a buffer with a pH of 2.532 ?

grams sodium nitrite = _____ g.

C. How many grams of solid ammonium bromide should be added to 1.50 L of a 0.171 M ammonia solution to prepare a buffer with a pH of 8.340 ?  

grams ammonium bromide = _____g.

Answer 1

5)

A.

PH= 2.374

HF= 2.000L of 0.212M

number of moles of HNO2= 0.212Mx2.00L= 0.424 moles

Ka of HF = 6.6x10^-4

Ka= 6.6x10^-4

-log[Ka]= -log(.6.6x10^-4)

PKa= 3.18

let be number of moles of KF = n moles

PH= Pka+log[salt/acid]

2.374 = 3.18 +log( n/0.424)

log( n/0.424) = - 0.806

n/0.424 = 10^-0.806

n/0.424 = 0.156

n= 0.156x0.424=0.0661 moles

number of moles of KF= 0.0661 moles

molar mass of KF= 58.1 g/mole

mass of 0.0661 moles of KF = 0.0661x58.1= 3.84 grams

mass of KF = 3.84 grams.

B)

PH= 2.532

HNO2= 0.500L of 0.154M

number of moles of HNO2= 0.154Mx 0.500L= 0.077 moles

Ka of HNO2 = 7.2x10^-4

-log[Ka] = -log(7.2x10^-4)

PKa= 3.14

let be number of moles of NaNO2 == n moles

PH= PKa + log[NaNO2/HNO2]

2.532 = 3.14 + log(n/0.077)

log(n/0.077) = - 0.608

n/0.077 = 10^-0.608

n/0.077 = 0.247

n= 0.247x0.077 =0.019

number of moles of NaNO2= 0.019 moles

molar mass of NaNO2= 68.99 g/mole

mass of 0.019 moles of NaNO2= 0.019 x 68.99 = 1.31 grams

mass of NaNO2= 1.31 grasm

C)

PH= 8.340

PH+POH=14

POH= 14-PH

POH= 14-8.340

POH= 5.66

Kb of NH3= 1.8x10^-5

-log[Kb] = -log(1.8x10^-5)

PKb = 4.74

NH3= 1.50L of 0.171M

number of moles of NH3= 0.171M x 1.50L=0.2565 moles

let be number of moles of NH4Br = n moles

POH = PKb + log[NH4Br/NH3]

5.66 = 4.74 + log(n/0.2565)

log(n/0.2565) = 0.92

n/0.2565 = 10^0.92

n/0.2565 = 8.318

n= 8.318x0.2565

n=2.133 moles

number of moles of NH4Br= 2.133 moles

Molar mass of NH4Br=97.94 gram/mole

mass of 97.94x2.133 = 308.91 grams

mass of NH4Br= 308.91 grams.