Help with CHEM. q, please? :)

[aquafreshlovin]

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Please show me how to work these problems? Have been working on them for awhile and still not getting it...

1) If 10mL of 1 M NaOH is titrated with 1 M HCl to a pH of 2, what volume of HCl was added? 😕 [answer is 10.2mL of HCl]

2) Which of the following combinations would produce a buffer solution of pH=4? (Ka HNO2 = 4.5e-4)

A)) 0.30M HNO2, 0.22M NaNO2
B)) 0.22M HNO2, 0.30M NaNO2
C)) 0.11M HNO2, 0.50M NaNO2
D)) 0.50M HNO2, 0.11M NaNO2

answer is C

THANKS!!!

 

[Mstoothlady2012]

2. First find the pka = -log Ka = 3.35
Then use H-H equation --> pH = pKa + log (base/acid)
4 = 3.35 + log (base/acid)
0.65 = log (base/acid)
antilog 0.65 = base/acid
4.5 = base/acid
Now use the answer choices to find which one gives you the ration of base:acid = 4.5 ---> 0.5/0.11 = 4.5 ---> C

 

[aquafreshlovin]

how do you do antilog 0.65 by hand? could u please show the steps? thanx!

 

[oohchki]

Please show me how to work these problems? Have been working on them for awhile and still not getting it...

1) If 10mL of 1 M NaOH is titrated with 1 M HCl to a pH of 2, what volume of HCl was added? 😕 [answer is 10.2mL of HCl]

For number 1:

Seeing that you are using a strong acid and base, that molarity is the same, and that pH is 2 (acidic), you know that there will be a greater volume of HCl than NaOH.

pH = 2 = -log(H3O+)
H3O+ = 10^-2 = 0.01 mol/L

H3O+ = (mol HCl - mol NaOH)/(vol. HCl + vol. NaOH)

0.01 mol/L = (1 mol/L x ? L HCl - 1 mol/L x 0.01 L NaOH)/(? L HCl + .01 L NaOH)
0.01 mol/L = (? mol - 0.01 mol)/(? L + 0.01 L)
0.01 ? mol + 0.0001 mol = ? mol - 0.01 mol
0.0101 mol + 0.01 ? mol = ? mol
0.0101 mol = 0.99 ? mol
0.0101 mol / 0.99 mol = ?
? = 0.01020202 L = ~10.2 mL

This question really stumped me for awhile. Hope you can follow my steps.

 

[oohchki]

how do you do antilog 0.65 by hand? could u please show the steps? thanx!

0.65 = log (x)

Remember:
c = log_ba
b^c=a

We're using log base 10, so b = 10.

0.65 = log₁₀x

so, since b^c=a...
10^0.65 = ~4.5

 

[nt4reall]

10mL of 1M NaOH will neutralize 10mL of 1M HCl at pH=7 . At that time , the solution is 20 mL of 1M HCl and 1M of NaOH. In order to have the solution which pH=2 , we have to add more acid ( HCl) .
CiVi= CfVf
Ci= 1M
Vi= 20 mL
Cf= 0.01 M ( because pH=2)
Vf= (CiVi)/Cf
= (1M*20 mL)/0.01M = 0.2 mL
We add 10ml of 1M Hcl to neutralize 10mL of 1M NaOH at pH=7 , then we need to add 0.2 mL HCl to get the solution which has pH=2 . Total volume of Hcl we add is 10mL+0.2 mL = 10.2 mL

 

[doc toothache]

Since NaOH and HCl have the same molarity, and since after neutralization the total volume will 20 ml the question becomes what volume of 1m HCl is needed to have 20 ml of 0.01m solution.

M1V1=M2V2
20 (0.01)=1V2
V2=0.2
Then total volume HCl is 10+ 0.2= 10.2

 

[perfectgrill]

.

 

[oohchki]

I understand the overall process, but I was wondering how you knew that 10^.65 = 4.5 without a calculator? I know 10^.5 is a little over 3, but how do we figure out 10^.65?

Oh, yikes, I'm not really sure... 😕

 

[aquafreshlovin]

hopefully DAT won't have a problem involving antilog...? 😕