pH 10.00인 NH3 완충용액 만들기. 0.200 M buffer 100 mL

pH 10.00인 NH3 완충용액 만들기. 0.200 M buffer 100 mL

 

 

Calculate the volume (mL) of concentrated ammonia (14.8 M of NH3) and

the weight (g) of ammonium chloride (NH4Cl, 53.5 g/mol)

you would have to take to prepare 100 mL of a buffer at pH 10.00

if the total concentration of buffer is to be 0.200 M.

 

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> NH3의 Kb = 1.80×10^(-5)

( pKb = 4.7447 )

 

 

 

pOH = 14 – pH = 14 – 10.00 = 4

 

 

 

pOH = pKb + log([짝산]/[약염기])

( 참고 https://ywpop.tistory.com/1926 )

 

4 = –log(1.80×10^(-5)) + log([짝산]/[약염기])

 

log([짝산]/[약염기]) = 4 + log(1.80×10^(-5)) = –0.7447275

 

[짝산]/[약염기] = 10^(-0.7447275) = 0.18

---> NH3 : NH4Cl = 1 : 0.18

 

 

 

pH 10.00 0.200 M NH3 완충용액 100 mL

(0.200 mol/L) (100/1000 L) = 0.02 mol

---> NH3 + NH4Cl = 0.02 mol

 

 

 

NH3의 mol수를 x라 두면,

NH3 : NH4Cl = 1 : 0.18 = x : (0.02–x)

0.18x = 0.02 – x

1.18x = 0.02

 

x = 0.02 / 1.18 = 0.016949 mol NH3

 

0.02 – 0.016949 = 0.003051 mol NH4Cl

 

 

 

[검산]

pOH = –log(1.80×10^(-5)) + log(0.003051 / 0.016949) = 4

 

 

 

0.016949 mol / (14.8 mol/L) = 0.001145 L = 1.145 mL NH3

 

0.003051 mol × (53.5 g/mol) = 0.1632 g NH4Cl

 

 

 

 

[ 관련 글 https://ywpop.tistory.com/17077 ] pH 8인 1.00 M NH3 완충용액 100 mL 만들기

 

 

 

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