Calculating pH

Chemistry Level 2

Ιn 200 m l 200ml aqueous solution ( S 1 ) (S1) of ethane acid ( C H X 3 − C O O H \ce{CH3-COOH} ) concentration C 1 = 0 , 25 M C1=0,25M ,dissolve 0,6g metallic M g \ce{Mg} without change of volume ,resulting solution S 2 S2 .

  • From the solution S 2 S2 remove 100ml and then diluted with water to a final volume of 250 m l 250ml find the p H pH of this solution S 3 S3

Use to the problem :

  • C H X 3 − C O O H \ce{CH3-COOH} : K a = 1 0 − 5 \ce{ Ka}=10^{-5}

  • H X 2 O \ce{H2O} : K w = 1 0 − 14 \ce{ Kw}=10^{-14}

  • M g \ce{Mg} : A r = 24 Ar=24

  • 0 , 1 − X = 0 , 1 0,1-X=0,1 if x x is very small.


The answer is 9.

Dimitris Kouroupos by dimitris kouroupos , 23, Greece

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1 solution

  • For metallic M g \ce{Mg} : n n = m M r \frac{m}{Mr} = 0 , 6 24 \frac{0,6}{24} = 0 , 025 m o l 0,025mol

  • In solution S 1 S1 for C H X 3 − C O O H \ce{CH3-COOH} : n = C 1 ∗ V 1 = 0 , 25 M ∗ 0 , 2 L = 0 , 05 m o l n=C1*V1=0,25M*0,2L=0,05mol

we make the reaction

M g \ce{Mg} ¨+ C H X 3 − C O O H \ce{CH3-COOH} => ( C H X 3 C O O ) X 2 M g \ce{(CH3COO)2Mg} + H X 2 \ce{H2}

0,025mol 0,05mol --------------- --------------
-0,025mol -0,05mol 0,025mol 0,025mol
-------------- --------------- 0,025mol 0,025mol
  • in solution S 2 S2 with volume V 2 = 200 m l V2=200ml we remove 100ml
200ml solution 0,025mol ( C H X 3 C O O ) X 2 M g \ce{(CH3COO)2Mg}
100ml x

x = 0 , 0125 m o l x=0,0125mol ( C H X 3 C O O ) X 2 M g \ce{(CH3COO)2Mg}

  • then diluted with water to a final volume of 250 m l 250ml

C 3 = C3= 0 , 0125 m o l 0 , 25 L \frac{0,0125mol}{0,25L} = 0 , 05 M 0,05M for ( C H X 3 C O O ) X 2 M g \ce{(CH3COO)2Mg} in solution S 3 S3

  • IN SOLUTION S 3 S3 :

( C H X 3 C O O ) X 2 M g \ce{(CH3COO)2Mg} => 2 C H X 3 C O O X − \ce{CH3COO^{-}} + M g X 2 + \ce{Mg^{2+}}

0,05M 0,1M 0,05M

FROM C H X 3 C O O X − \ce{CH3COO^{-}} :

C H X 3 C O O X − \ce{CH3COO^{-}} + H X 2 O \ce{H2O} => C H X 3 − C O O H \ce{CH3-COOH} + O H X − \ce{OH^{-}}

(0,1-X)M ----------------- X X
  • K a = 1 0 − 5 \ce{ Ka}=10^{-5} and Beucause K w = 1 0 − 14 \ce{ Kw}=10^{-14}

K a \ce{ Ka} * K b = \ce{ Kb}= K w \ce{ Kw} => K b = 1 0 − 9 \ce{ Kb}=10^{-9} for C H X 3 C O O X − \ce{CH3COO^{-}}

believed to be around 0,1-Χ=0,1 BECAUSE X IS VERY SMALL K b = \ce{ Kb}= X 2 ( 0 , 1 − X ) \frac{X^2}{(0,1-X)} = X 2 ( 0 , 1 ) \frac{X^2}{(0,1)} => 1 0 − 10 = X 2 10^{-10}=X^2 => X = 1 0 − 5 X=10^{-5}

this means

p O H = − l n ( 1 0 − 5 ) = 5 pOH=-ln(10^{-5})=5 ,because p O H + p H = 14 pOH+pH=14 => p H = 14 − 5 = 9 pH=14-5=9

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