A saturated solution of magnesium hydroxide the filtered solution is placed in a

Question

A saturated solution of magnesium hydroxide the filtered solution is placed in a flask with 25 ml of deionized water and 3 drops of phenolphthn. is titrated with 2.56 x 10 M hydrochloric acid until the solution in the flask turns clear. If the inal buret reading is 10.16 ml and the final buret reading at the equivalence point is 2228 mil, e it Kap of magnesium hydroxide. What is its solubility? What would you predict its was dissolved in 0.3 M NaOH (you will want to use an ICE table to work this one is filtered until the solution is clear. A 10.00 ml aliqust of solubility to be if it out)? 6. Titration Plot of a Weak Acid with a Strong Base 13 12 10 pH Above is the titration plot of a weak acid with a strong base. To the nearest tenth, what is the pka of the acid? What is the Ka of the acid? If the literature value of the Ka is 1.0 x 10, what is the percent error in Ka? What is the percent error in pKa?

Explanation / Answer

1) titration of saturated solution of Mg(OH)2 with HCl:

volume of HCl used up in titration=final burette reading -initial burette=22.28ml-10.16ml=12.12ml

mol of HCl used up=molarity *volume=(2.56*10^-4M)*0.01212L=3.103*10^-6 mol

Mg(OH)2 +2HCl --->MgCl2 +2H2O

mol Mg(OH)2 /mol HCl=1/2

So,mol of Mg(OH)2=1/2*mol HCl=1/2*(3.103*10^-6 mol)=1.551*10^-6 mol

[Mg(OH)2]diluted=concentration =mol/volume=(1.551*10^-6 mol)/35ml=(1.551*10^-6 mol)/0.035L=4.431*10^-5 M

[Mg(OH)2]saturated=[Mg(OH)2]diluted* volume of diluted

So,[Mg(OH)2]saturated=[Mg(OH)2]diluted*( volume of diluted/volume of saturated)=(4.431*10^-5 M)*(35ml/10ml)=1.551*10^-4 M

Mg(OH)2 (s) <---->Mg2+(aq)+2OH-(aq)

[Mg(OH)2]saturated=[Mg2+] =1.551*10^-4 M

[OH-]=2*[Mg(OH)2]saturated=2*(1.551*10^-4 M)=3.102*10^-4 M

Ksp=[Mg2+][OH-]^2=(1.551*10^-4 M)(3.102*10^-4 M)^2=2.315*10^-15

Ksp of Mg(OH)2=2.315*10^-15

solubility =[Mg(OH)2]saturated=1.551*10^-4 M

[OH-]=0.3M

Using ICE table :

Ksp=[Mg2+][OH-]^2

or,(2.315*10^-15)=x*(0.3+x) [but x<<<0.3M)

or, (2.315*10^-15)=x*(0.3)

x=[Mg2+]=7.716*10^-15M

Thus, solubility=[Mg2+]=7.716*10^-15M

[Mg(OH)2](s) [not included in eqm expression) [Mg2+] [OH-] initial 1.551*10^-4 M 0 0.3M change -x +x +x equilibrium 1.551*10^-4 M -x x 0.3+x

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