Calculating pH

[RHONDAROBINSON]

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I have two pH problems that I need help with.


1. Wha tis the pH of a .05M of NH3 solution if Kb is 1.8Xe-5.

2.If 30ml of a .02M HBr solution is mixed with 40ml of .03 KOH, what is pH of the mixture.

Rhonda

 

[Lindbergjb]

1. pH= 11.4772
nh3 + h20 = nh4 + oh

Kb=nh4 x oh / .5

oh=.003 -log(.003) =2.522 14-2.522= 11.4772

 

[Lindbergjb]

2.

HBr+OH = Br + H2O

HBR= 30mL x 0.2M = 6moles
KOH= 40mL x 0.3M = 12moles

pOH= -log( 6moles OH/ 40mL KOH) = 0.823

14 - 0.823 = pH = 13.176

 

[doc toothache]

2.
HBr+OH = Br + H2O
HBR= 30mL x 0.2M = 6moles
KOH= 40mL x 0.3M = 12moles
pOH= -log( 6moles OH/ 40mL KOH) = 0.823
14 - 0.823 = pH = 13.176

The total volume of the solution will be 70 ml.

 

[RHONDAROBINSON]

I do apologize for not putting the answer to these questions. I need to know an explanation on how the book arrived at these answers. Doc Tootache and Lindergjb thanks for your help.

Questions:

What is the pH of a .05M NH3 solution if Kb is 1.8E-5? Answer is 11.

If 30ml of .02M HBr solution is mixed with 40ml of .03M KOH, what is the pH of the mixture? ANSWER IS 12.

Rhonda

 

[Lindbergjb]

oh yea, my bad...it was late

 

[doc toothache]

What is the pH of a .05M NH3 solution if Kb is 1.8E-5? Answer is 11.

If 30ml of .02M HBr solution is mixed with 40ml of .03M KOH, what is the pH of the mixture? ANSWER IS 12.
Rhonda

Kb=1.8 E-5= [NH4+][OH-]/NH4OH = 1.6E-5=X E2/0.5
X2= 90 x 10 E8; [OH-]= 9.48 x 10 E4; pOH= 3.13, pH= 10.87

30 x .02= 6
40 x .03=12
Therefore, 20ml of 0.3M left after neutralization when the final volume will be 70ml.

V1M1=V2M2
20 x 0.03= 70M2; M2=8.57 x 10 E3; pOH= 2.1, pH =11.9

 

[illbirz]

1. pH= 11.4772
nh3 + h20 = nh4 + oh

Kb=nh4 x oh / .5

oh=.003 -log(.003) =2.522 14-2.522= 11.4772

How do we know that OH is 0.003? 😕

 

[iwannabadentist]

Kb=1.8 E-5= [NH4+][OH-]/NH4OH = 1.6E-5=X E2/0.5
X2= 90 x 10 E8; [OH-]= 9.48 x 10 E4; pOH= 3.13, pH= 10.87

30 x .02= 6
40 x .03=12
Therefore, 20ml of 0.3M left after neutralization when the final volume will be 70ml.

V1M1=V2M2
20 x 0.02= 70M2; M2=8.57 x 10 E3; pOH= 2.1, pH =11.9

did you mean 20*0.03 = 70*M2?
Because the KOH was at a concentration of 0.03M rite?

 

[doc toothache]

did you mean 20*0.03 = 70*M2?
Because the KOH was at a concentration of 0.03M rite?

Yes. The appropriate correction was made.

 

[drdolittle1]

Could this type of question show up on the DAT?? I would think not...seems tough to calculate w/o a calculator..

 

[RHONDAROBINSON]

This question was one in the DAT Destroyer.


Rhonda

 

[jbrenn4]

I actually have a diff way it may be easier ...it just deals with scientific notation,,
NH3 + H2) -> NH4+ + OH-

Kb= [NH4+][OH-]/[NH3]

1.8 E5- just round to 2 E -5

2 E-5= x2/5e-2

10e-7 x2..move your decimal places to get 100e-8 so you can easily take the sqaure root.....10 e-4....which is really 1e-3....pOH = 3 therefore pH = 11