chem qes

[zvm77]

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hi

I encountered this qes, how would I go about getting the ans

"what volume of HCl was added if 20 ml of 1M NaOH is titrated with 1M HCl to produce a ph=2"

Thank you in advance.

 

[rdhdds1]

hi

I encountered this qes, how would I go about getting the ans

"what volume of HCl was added if 20 ml of 1M NaOH is titrated with 1M HCl to produce a ph=2"

Thank you in advance.

m1v1=m2v2

 

[zvm77]

m1v1=m2v2

This formula is used to find 1 vol based on another vol using this formula I'd get an hcl vol of 20ml, now 20 ml of strong base + 20 ml of strong acid of the same molarity would give me a neutral salt, but the qes states that the resulting soln would have a ph of 2, therefore, how much acid (much more than initial base) must be added to get it to a ph of 2?

 

[ucla2134]

hi

I encountered this qes, how would I go about getting the ans

"what volume of HCl was added if 20 ml of 1M NaOH is titrated with 1M HCl to produce a ph=2"

Thank you in advance.

There is two part of the problem that need to take into account.
Titration and dilution which have totally different formular
Titration: n1v1=n2v2 (n is for Normality)
Dilution: M1V1=M2V2

The first part of the problem is for the titration (base titrated with acid to yield the equivalence point)
n1v1=n2v2--> 20*1=1*v2--> thus need 20 ml of HCL to yield the equivalence point.
So at the equivalence point you have: 20 ml OH- + 20 ml H+. Total is 40 ml at the equivalence point.
After the equivalence point, the extra amount of HCL added will determined the pH of the solution---> DILUTION
M1V1=M2V2
1M*x=0.02(40+x). x is the volume amount of HCL added after the equivalence point.
solve for x, you got 0.8. Thus the final volume is 40.8, thus need 20.8L of HCL.

 

[JBarr29]

i was just typing that haha if you need more help check out http://en.allexperts.com/q/Chemistry-including-Biochemistry-1355/2008/8/Titration-4.htm
for another problem/more practice

 

[Danny289]

There is two part of the problem that need to take into account.
Titration and dilution which have totally different formular
Titration: n1v1=n2v2 (n is for Normality)
Dilution: M1V1=M2V2

The first part of the problem is for the titration (base titrated with acid to yield the equivalence point)
n1v1=n2v2--> 20*1=1*v2--> thus need 20 ml of HCL to yield the equivalence point.
So at the equivalence point you have: 20 ml OH- + 20 ml H+. Total is 40 ml at the equivalence point.
After the equivalence point, the extra amount of HCL added will determined the pH of the solution---> DILUTION
M1V1=M2V2
1M*x=0.02(40+x). x is the volume amount of HCL added after the equivalence point.
solve for x, you got 0.8. Thus the final volume is 40.8, thus need 20.8L of HCL.

you are right just remeber pH=2 =====> [H+]= 0.01 not 0.02 and tit ends to~ 20.4 ml we need.

 

[klutzy1987]

A little while back this problem was brought up and it was explained in great detail. Run a search and see if you can find it.