- #1

Conductivity

- 87

- 4

For example let's say, If we add 200 mL of NaOH to 400 mL of HNO2 2M, The pH become 1.5 + (##pH_i##)

calculate the concentration of NaOH.

The usual way of doing is this:

## K_a = (0.02+x)[H^+]/(4/3-x) ## then find the x which most of the time is equal to the initial concentration of OH- after addition which you can convert back to the original solution and find the required information.

But the way I see how these approximations work is that you have to also think about how the additional OH can interact with H+ (or H3O+ if you like) to form water. So you really need simultaneous equations

##K_w = ([H+]_i -y)([OH-]_i-x-y) ##

Where X is from the original acid reaction and Y comes from forming water reaction.

Now usually Y is negligible so I can assume the OH- fully reacts with the acid but here in this situation Y isn't

I did the math and found that x = 0.41345 and y = 0.01917811 which is not negligible ( about 4.63% mistake, However I am talking about more extreme version of this) . Is this way of thinking is true?