Gchem

[Glycogen]

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Determine the pH of a 1.3*10^-5M solution of H3BO3(ka=5.4*10^-10)

a.2.4
b.4.8
c.5.2
d.6.4
e.7.1

Thanks.
By the way,I have the answer, I just need explanation.

 

[osimsDDS]

Determine the pH of a 1.3*10^-5M solution of H3BO3(ka=5.4*10^-10)

a.2.4
b.4.8
c.5.2
d.6.4
e.7.1

Thanks.
By the way,I have the answer, I just need explanation.

Is the answer E, 7.1 ???

 

[Glycogen]

Yes it is, but why?
I was wonder where you are!

 

[osimsDDS]

haha sorry i went to go eat at this nice Indian buffet place haha i feel really tired tho, indian food is heavy and makes me sleepy hahaha

Anyways back to the question:

they give you Ka, and an Acid (H3BO3) so it will dissociate as follows:

H3BO3 ---> H+ + H2BO3-

Now figure out your Ka formula which is products over reactants...

Ka = [H+][H2BO3-]/[H3BO3]

Now just plug in the knowns vs unknowns:

5.4*10^-10 = x^2/(1.3*10^-5M)
x = 8.4 x 10^-8, this is your H+ concentration and also H2BO3- concentration because they both equal to x

So now take the H+ concentration and find pH because the negative log of Ka is the pH

pH= 8 - log 8 = 7.1

Basically log 10 = 1 and log 1 = 0 so log 8 will be closer to 1 than zero...

Hope that helps

 

[excusezmoi]

Determine the pH of a 1.3*10^-5M solution of H3BO3(ka=5.4*10^-10)

a.2.4
b.4.8
c.5.2
d.6.4
e.7.1

Thanks.
By the way,I have the answer, I just need explanation.

[HA] --> [H+] + [A-]
ka= [H][A] / [HA]
5.4*10^-10 * 1.3*10^-5 = x^2
7.0 *10^-15 = x^2
- log (8.37 * 10^- 8) = [H+]
7.07 = pH ~ E

 

[Glycogen]

Thank you so much.See I have done all that and then I end up with a completely acidic solution.The reason was that,I did not pay attention that I have to take out of radical.
I was so disappointed to miss such a easy Q!
I feel better now ...🙂

 

[Andrew324]

I got the formula and everything right on the calulator, but how do you guys do that first step

5.4*10^-10 * 1.3*10^-5
on paper.

Multiply the numbers and add the sci notation.

5.4*1.3= apx 7
and add the sci notation -10 + -5 = -15

Is that how you guys do it?

 

[akirar]

haha sorry i went to go eat at this nice Indian buffet place haha i feel really tired tho, indian food is heavy and makes me sleepy hahaha

Anyways back to the question:

they give you Ka, and an Acid (H3BO3) so it will dissociate as follows:

H3BO3 ---> H+ + H2BO3-

Now figure out your Ka formula which is products over reactants...

Ka = [H+][H2BO3-]/[H3BO3]

Now just plug in the knowns vs unknowns:

5.4*10^-10 = x^2/(1.3*10^-5M)
x = 8.4 x 10^-8, this is your H+ concentration and also H2BO3- concentration because they both equal to x

So now take the H+ concentration and find pH because the negative log of Ka is the pH

pH= 8 - log 8 = 7.1

Basically log 10 = 1 and log 1 = 0 so log 8 will be closer to 1 than zero...

Hope that helps

How do you do this step without a calculator?

Thanks

 

[osimsDDS]

How do you do this step without a calculator?

Thanks

first of all the DAT will never give you anything that you cant do mathematically without a calculator, usually they give you nice round numbers...but stupid test prep people like kaplan give you long really annoying numbers that you HAVE to use a calculator or maybe just round...if you get to that point where the numbers are weird just round them and cross your fingers..

 

[Glycogen]

The was I do,I multiply 5.4*1.3=7 and the add up the powers-->-10-5=-15
now you have 7*10^-15=x^2
x= 9*10^-8
Which almost the same as 8.4 *10^-8

 

[HoangDDS]

I got the formula and everything right on the calulator, but how do you guys do that first step

5.4*10^-10 * 1.3*10^-5
on paper.

Multiply the numbers and add the sci notation.

5.4*1.3= apx 7
and add the sci notation -10 + -5 = -15

Is that how you guys do it?

I just multiplied it out. 5.4 x 1.3 = 6.02 which i rounded to 6. Add the exponents = 6 x 10^-15 = x^2

x = sqrt( 60 x 10^-16)
= 4sqrt15 x 10^-8

x = [H+] so find a choice with a pH in the 7 range. E is the closest option.

 

[Andrew324]

Determine the pH of a 1.3*10^-5M solution of H3BO3(ka=5.4*10^-10)

Just for for how about we change the Question to say

Determine the pH of a 1.3*10^-5M solution of H3BO3(kb=5.4*10^-10)

A) pH 3.2
B) pH 5.4
C) pH 6.9
D) pH 7.6
E) pH 9.1

 

[I-Baby]

haha sorry i went to go eat at this nice Indian buffet place haha i feel really tired tho, indian food is heavy and makes me sleepy hahaha

Anyways back to the question:

they give you Ka, and an Acid (H3BO3) so it will dissociate as follows:

H3BO3 ---> H+ + H2BO3-

Now figure out your Ka formula which is products over reactants...

Ka = [H+][H2BO3-]/[H3BO3]

Now just plug in the knowns vs unknowns:

5.4*10^-10 = x^2/(1.3*10^-5M)
x = 8.4 x 10^-8, this is your H+ concentration and also H2BO3- concentration because they both equal to x

So now take the H+ concentration and find pH because the negative log of Ka is the pH

pH= 8 - log 8 = 7.1

Basically log 10 = 1 and log 1 = 0 so log 8 will be closer to 1 than zero...

Hope that helps

Dude you just saved my life lol. I'm here freaking out trying to do destroyer problems with no calc and doing it like this super easy. Thanks!

 

[klutzy1987]

I just multiplied it out. 5.4 x 1.3 = 6.02 which i rounded to 6. Add the exponents = 6 x 10^-15 = x^2

x = sqrt( 60 x 10^-16)
= 4sqrt15 x 10^-8

x = [H+] so find a choice with a pH in the 7 range. E is the closest option.

5.4*1.3 is actually 7.02

To do that without a calculator, just reason that 2 *5.4 is 10.8 and 1*5.4 is 5.4.
Therefore, 1.3 *5.4 is about a third of the way between 5.4 and 10.8. That is approximately 7.