Masa kalijum laktata

Submitted by MD on Sun, 12/22/2013 - 17:13
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Koliko grama K-laktata, Mr = 128.2 g/mol, treba dodati u 750cm^3 rastvora monobazne mlečne kiseline koncentracije 0,05 mol/dm^3 da bi se dobio puferski rastvor čija pH iznosi 4,1? Kk mlečne kiseline iznosi 1,4*10^-4 mol/dm^3.

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milicaradenkovic - Sun, 12/22/2013 - 17:58

CH3CH(OH)COOH-------------->CH3CH(OH)COO- + H+

pH=4.1 pa je{H+}=10(na -pH)=7.94*10(-5) mol/l

Kk={CH3CH(OH)COO-}*{H+}/{CH3CH(OH)COOH}

1.4*10(-4)={CH3CH(OH)COO-}*7.94*10(-5)/0.05

{CH3CH(OH)COO-}*7.94*10(-5)=7*10(-*7)

{CH3CH(OH)COO-}=7*10(-7)/7.94*10(-5)=0.0088 mol/l

CH3CH(OH)COOK--------->CH3CH(OH)COO- + K+

{CH3CH(OH)COOK}={CH3CH(OH)COO-}=0.0088 mol/l

n(CH3CH(OH)COOK)={CH3CH(OH)COOK}*V=0.0088*0.75=0.0066 mol

m(CH3CH(OH)COOK)=m(CH3CH(OH)COOK)*Mr=0.0066*128.2=0.846 g