Chemistry pKa problem?

Question

25 mL of a 0.075 M solution of a weak acid was titrated with 0.10 M NaOH. After the addition of 5.0 mL of NaOH the pH was 4.85.
a. What is the pKa of the weak acid?
b. If the solution at this point is diluted to 50 mL for use as a buffer, what is the
molar concentration of the buffer?

Answer 1

I assume that they mean a weak monoprotic acid (1 H+ per acid molecule). Thus the acid can be written as HA.

For simplicity we can use the Henderson-Hasselbach equation

pH=pKa+log[A-]/[HA] (1)

[A-] comes from the complete reaction of NaOH with HA. Thus mole A- = mole NaOH and [A-]= moleNaOH/Vtotal = M(NaOH)*V(NaOH)/ Vtotal = 0.1*5/(25+5) =0.5/30

[HA]= (mole HA(total) -mole HA(reacted)) / Vtotal
The reaction of the acid with NaOH is 1:1 thus moleHA(reacted)=mole NaOH and thus
[HA]= (0.075*25-0.1*5)/30 =1.375/30

Note that I didn't convert mL into L since in the ratio the convertion factor is simplified. If you want to be exact you need to multiply each volume value with 10^-3.

Using (1)
4.78=pKa+log(0.5/1.375) =>
pKa= 4.78-log(0.5/1.375) =5.22

Note that this solution inherently does some approximations/ assumptions. You can get a more exact solution if you use a suitable ICE table and solve the quadratic equation. If you need that type of solution as well let me know.



The molar concentration of the buffer will be the molar concentration of A in total (both as HA and A-). But all A comes from the original HA. So finding the molar concentration of the buffer is the same as determining the molar concentration of HA when diluting the initial solution of the weak acid from 25 mL to 50 mL.
C1V1=C2V2 =>
C2=C1V1/V2= 0.075*25/50= 0.0375 M

Answer 2

concentration of acid = [acid] = 0.015M concentration of salt = [salt] = 0.035M pH = 4.fifty six pka=? pH = pKa + ( [salt] / [acid]) resolve it your self , i'm hoping ur stable at maths.