Neutralization - Strong polyprotic acid and strong base

[alebeau]

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Hi,

Can anyone tell me if I have equal molar concentrations of a strong polyprotic acid and strong base, will the resulting pH be 7?

For instance, in the following equation, H2SO4 + NaOH --> Na+ + HSO4- + H2O, if I started off with .1 M H2SO4 and .1 M NaOH, would the pH be 7?

Thanks,

AL

 

[Teleologist]

Hi,

Can anyone tell me if I have equal molar concentrations of a strong polyprotic acid and strong base, will the resulting pH be 7?

Bruh, the pH ain't 7.

Lmk if you want to ask anything else.

 

[alebeau]

Thanks Teleologist for your quick response. If I might just ask you two quick follow up questions,

1) If I had .1 M HCl and .1 M NaOH, the pH would be 7. Why would this not be the case for H2SO4 and NaOH ?

2) In regard to the question above, could you please tell me how to work out mathematically what the pH would be if I had .1 M H2SO4 and .1 M NaOH ?

Thanks bud,

AL

 

[type12]

You need the ka of HSO4- before you can figure it out.

 

[alebeau]

Ah, I see where you're going.

Tell me if I am correct in my reasoning:

H2SO4 + NaOH -- > HSO4- + H2O
I .1 .1 0 N/A
C -.1 -.1 +.1 N/A
E 0 0 .1 N/A

HSO4- + H2O -- > SO42- + H+ Ka = 1.2 x 10^-2
I .1 N/A 0 0
C -x N/A +x +x
E .1-x N/A x x

Q = Ka
([H+][SO42-])/([HSO4-) = 1.2 x 10^-2
x^2/(.1-x) = 1.2 x 10^-2
x = 0.029156791662493894

x = [H+] ==> pH = -log(0.029156791662493894) = 1.53

Is this correct? If it is, just one last thing, how would one go about figuring out how much base is needed to completely neutralize the solution? Thanks

(PS: I apologize beforehand for any formatting problems - everything appeared fine on my end)

 

[Teleologist]

Thanks Teleologist for your quick response. If I might just ask you two quick follow up questions,

1) If I had .1 M HCl and .1 M NaOH, the pH would be 7. Why would this not be the case for H2SO4 and NaOH ?

2) In regard to the question above, could you please tell me how to work out mathematically what the pH would be if I had .1 M H2SO4 and .1 M NaOH ?

Thanks bud,

AL

1) If I give you a box with two cats inside and then I give you a box with one cat inside, how come one box has twice the number of paws as the other box?

2) You would need the Ka of hydrogen sulfate anion. Fortunately I know what it is: 1.2*10^-2.

 

[Teleologist]

Is this correct? If it is, just one last thing, how would one go about figuring out how much base is needed to completely neutralize the solution? Thanks

(PS: I apologize beforehand for any formatting problems - everything appeared fine on my end)

How many equivalents of base would you need to neutralize the acid?

 

[alebeau]

My guess is two because it's H2SO4. Is that two say that .2 NaOH could neutralize .1 M H2SO4? Thanks

 

[type12]

Ah, I see where you're going.

Tell me if I am correct in my reasoning:

H2SO4 + NaOH -- > HSO4- + H2O
I .1 .1 0 N/A
C -.1 -.1 +.1 N/A
E 0 0 .1 N/A

HSO4- + H2O -- > SO42- + H+ Ka = 1.2 x 10^-2
I .1 N/A 0 0
C -x N/A +x +x
E .1-x N/A x x

Q = Ka
([H+][SO42-])/([HSO4-) = 1.2 x 10^-2
x^2/(.1-x) = 1.2 x 10^-2
x = 0.029156791662493894

x = [H+] ==> pH = -log(0.029156791662493894) = 1.53

Is this correct? If it is, just one last thing, how would one go about figuring out how much base is needed to completely neutralize the solution? Thanks

(PS: I apologize beforehand for any formatting problems - everything appeared fine on my end)

The math is correct, but your answer is too close to 0.1 to do the MCAT assumption that x is really small. So, you cannot solve for x without quadratics.

How can you tell beforehand? Well, the ka seems high, as in, the desire for this weak acid to disassociate is too high.

The MCAT would not ask you a question where x will be a significant value when subtracting from the original concentration.

How much base is needed to neutralize a weak acid or base is beyond the scope of the MCAT. You need to know only whether the equivalence point will be less than or greater than 7. Why? Because weak acids are in equilibrium with their acid and conjugate forms (don't disassociate completely). You have to coax it to give up its H+'s, by adding more base. Besides that, it's simple math, but more than the MCAT requires, and even the term "neutralize" is sometimes discomforting to use in this scenario (weak acid).

 

[alebeau]

Wonderful. Thanks for that answer type12.

I deeply appreciate it.

ADL