pka vs ka

[Rayna Irving 2C]

Why is it that we can add pka + pkb to get pkw but we have to multiply ka and kb to get kw?

[John Antowan 1K]

Since pka usually deals with small decimals, adding the values will not get to the desired value of 1x10^-14. The only way to achieve this with decimals is by multiplying.

[Stuti Pradhan 2J]

I believe it is because of the product property of logarithms which states that
log(MN) = log(M) + log(N). Multiplying Ka and Kb will give you Kw, but if you want to find the pKw, you would be doing log(Kw)=log(Ka*Kb), which equals log(Ka)+log(Kb) or pKa+pKb.

Hope this helps!

[Randi Ruelas 1G]

The values for Ka and Kb are small, and when we multiply numbers with exponents we would just add the exponents, to give us an even smaller number. It makes sense to add pka and pkb due to the fact, it is the -log for ka and kb, so it is giving us a much greater number

[John_Tran_3J]

Since it p values involves -log, you would be given valets that you have to multiply or divide.

[Nina Ellefsen 2D]

Since pKa and pKb are both -log values with the same base, saying pKa+pKb is the same as saying -log(Ka*Kb)

[TiffanyBrownfield 2I]

When you deal with exponents, you know that when you divide or multiply values with exponents, you simply add or subtract them. (ex. 10^-14 / 10^-7 = 10^-7). The pKa takes the -log base 10 of the Ka value, so it's basically taking on the value of the exponent in a way. This is why you add when you deal with pKa and multiply with kA.

[Sarah Salam 1J]

You add pKa and pKb because they are logarithmic values, and based on their properties adding them is like multiplying Ka and Kb.

[Luveia Pangilinan 1A]

this is because that is a smaller value compared to the other two :D

[Arnav Saud 2C]

If I'm correct, the reason why we can add pka and pkb to get pkw is because pka and pkb are found by taking the logmarithm of the ka and kb. When you multiply two logs together, you can add their inner parts up.
Thus,
-log(ka*kb)) = -log(ka) + -log(kb) ---> This should translate to pka + pkb

[MariaCassol1L]

This is because of log laws.
If we assume that Ka x Kb = Kw is true

we take the -log of both sides:
-log(Ka x Kb) = -log (Kw)

log law: log(ab)=log(a) + log(b):
-log(Ka)+ (-log(Kb))= -log(Kw)

So since pKa=-logKa, pKb=-logKb, pKw=-logKw you can write:
pKa+pKb=pKw

I believe Dr.Lavelle has some math review on the 14B website, but just to add on a similar law applies to division:
log(a/b)= log(a)-log(b)

[Bai Rong Lin 2K]

I remember something about it being a smaller value.

[Karl Yost 1L]

It's because of log properties:

logAB= logA + logB

[Britney Tran IJ]

you're adding pKa and pKb because they're log functions

[Kathy_Li_1H]

Hi! I believe the reasoning behind this is the logarithm function for pka (logab = log a + log b).

[Lung Sheng Liang 3J]

Hello, it's because pKa and pKb are the log functions of Ka and Kb

[Victor Qiu 1C]

log₁₀(ab) = log₁₀(a) + log₁₀(b) is correct

Also,
if pKw = pKa + pKb
10^-pKw = 10^-(pKa+pKb)
10^-pKw = 10^-(pKa) × 10^-(pKb)
Kw = Ka × Kb

[Morgan Gee 3B]

pKa is the -log[ka]. This changes the relationships because of the way logs work. When you do 10^-14 = ka x kb and take the negative log of that equation, you end up with 14 = pKa + pKb

[Astha Sahoo 3I]

We use the rule for logs. Ln(a) + ln(b) = ln(a*b). Thus, we multiply the values for concentrations, but for the pKa and pKb, since they are logs, they use the rule of logs.

[David Y]

logAB= logA + logB

[Jose Miguel Conste 3H]

its because the log rules, log(pka) + log(pkb) = pka*pkb

[Brian Acevedo 2E]

This has to do with the logarithm rule which states that the sum of two logs (let's call them logx and logy) is equal to the log of the product of the numbers you were taking the log of previously. (i.e. log(x*y))

[Teti Omilana 1G]

since p=(-log), you can use the log rules: log(pka)+log(pkb)=pka*pkb

[Neha Gupta 2A]

I think it's because for pH and pKa you are able to add pH and pOH to get 14 and pKa +pKb to get 14. When [H] and Ka are in their regular form you can't add them together to get 14 because you haven't taken the -log of them.

[George Hernandez 3I]

Its basically the same as adding pH and pOH to get 14. The logarithmic properties allow you to add the pKa and pKb values to get pKw. But the concentrations themselves must be multiplied.

[Pratika Nagpal]

what is the difference between pka and ka and pkb and kb

[Charisma Arreola 2F]

I think it’s because of the difference and how small pKa values typically are.

[Diana Aguilar 3H]

I think it has to do with log rules, logab = log a + log b, and also because pKa and pKb are log functions.

[Naomi Hernandez-Ramirez 1J]

Both pKa and pKb use -log values with the same base pKa+pKb is the same as saying -log(Ka*Kb)