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serg [7]
3 years ago
5

Calculate the ph of a buffer that is 0.13m in lactic acid and 0.10m in sodium lactate.

Chemistry
1 answer:
Alona [7]3 years ago
3 0
According to Henderson–Hasselbalch Equation,

                                       pH  =  pKa + log [Lactate] / [Lactic Acid]
As,
      Ka of Lactic Acid  =  1.38 × 10⁻⁴

           pKa  = -log Ka
           pKa  = -log 1.38 × 10⁻⁴
           pKa  =  3.86
So,
                                  pH  =  3.86 + log [0.10] / [0.13]

                                  pH  =  4.74 + log 0.769

                                  pH  =  4.74 - 0.11

                                  pH  =  4.63
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kompoz [17]

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

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V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

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So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

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For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

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V(standard 2) = 4.63 + 5.17

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M (FeSCN⁺²)  = 0.00109/9.8

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3 years ago
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Answer:

The correct answer is -

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