[Tanny]

How to find the pH of 32 mL of 0.1 M of NaOH added to 100mL of 0.05 of NaHCO3
Steps please!!!
OH-    +    HCO3 - --->    CO2    - +H2O

2.4x 10^-2      3.78 x 10^-2                   0

 -x        -x                   +x

 4.68 x 10^-11 = [2.4x 10^-2-x][3.78 x 10^-2-x]/

x = 2.06 x 10^-7

There are no H3O+ or OH -, how do I find the pH???

[Rutherford]

A NaHCO3 and Na2CO3 buffer is made. Use the Henderson-Hasselbach equation to calculate the pH.

[Tanny]

OH-    +    HCO3 - --->    CO2    - +H2O

pH = pKa + log (base/acid)


What is the base? Is it just the OH or multiplied the HCO3? Since HCO3 produces a conjugate CO2- acid, is that a base too??

I made a mistake up there. The x is x = 5.95 x 10 ^-3.
Is the equation 10.329 + log ([OH-][HCO3]-/[CO2]-)? or 10.329 + log ([OH-]/[CO2]-)??
I also need help finding the buffer capacity. I never learned it but I need to do it.
I have an one hour limit.

[Rutherford]

NaHCO3+NaOH-->Na2CO3+H2O
Calculate the concentration of the produced Na2CO3 and the concentration of the NaHCO3 that left in the solution. Na2CO3 will ionize to 2Na⁺ and CO3^2-, and NaHCO₃ will ionize to Na⁺ and HCO3^-. HCO3^- is the acid because it is a proton donor (it can donate a H⁺ ion) and CO3^- is the base because it can only accept a proton (H⁺). pKa you can calculate from the Ka (pKa=-logKa).

[Tanny]

What about the OH on NaOH? Is that ignored? I already found their respective concentrations?

[Rutherford]

NaOH reacts completely with NaHCO3 to form Na2CO3.