According to the information of the picture, calculate the \"predicted\" pH of s

Question

According to the information of the picture, calculate the "predicted" pH of solution # 2, #6 , #8 Calibri 12 AT General Assigned pH to make buffer 5 Name of weak acid assigned acetic Ka (acetic acid) 1.76 e-5 6 Name of conj. base assigned Sodium acetate 7 volume of 0.3M weak Acid used a volume of o.3M conj base used pH ApH (from Soln1) %error from assigned pH 11 Soln1 (20mL buffer) 12 Soln2 (20mL +1mL.5M HCI) 4.59 Y 13 Solna (20mL tlmL -5M NaoH) 4.91 Y o.15Y 14 Solna (20mL +1ml H20) 4.8 Y 0.04 Y 16 PH ApH (from Soln5) 17 Soln5 (20mL Distilled Water) 18 soln6 (20mL +1mL .5M HCI) 19 Soln7 (20mL +1m .5M NaOH) L12-21 YL 6.65 Y 21 Solna (20mL 0.3M acetic acid) 22 Soln9 (20mL o.3M acetate ion) 25 initial concentration weak Acid 26 initial concentration conj base 116 27 Predicted pH of soln w1 4.55

Explanation / Answer

Molarity of acid in the final buffer = 0.3/(100/61.31) = 0.184 M

Molarity of conjugate base (salt) in the final buffer = 0.3/(100/38.69) = 0.116 M

For solution 2:

This solution contains 20 mL of buffer and 1 mL of 0.5M HCl

Moles of acid present in 20 mL buffer = 0.02*0.184 = 0.00368

Moles of salt present in 20 mL buffer = 0.02*0.116 = 0.00232

Moles of HCl added = 0.001*0.5 = 0.0005

When these moles of acid are added, the net moles of acid increase while those of salt decrease by this same amount.

So in the Henderson Hasselbach equation:

pH = pKa + log(moles of salt/moles of acid)

Putting values we get:

pH = 4.75 + log((0.00232-0.0005) / (0.00368+0.0005)) = 4.39

For solution 6:

This solution contains 20 mL of distilled water and 1 mL of 0.5M HCl

So, moles of acid HCl present = 0.001*0.5 = 0.0005

Final volume = 20+1 = 21 mL = 0.021 L

So, [H+] = 0.0005/0.021 = 0.0238 M

So, pH = -log([H+]) = 1.623

For solution 8:

Let HA denote the acetic acid

The ICE table for the dissociation reaction is:

HA ----> H+ + A-

Initial 0.3 0 0

Eqb 0.3-x x x

K = x2/(0.3-x) = 1.76*10-5

Solving we get:
x = 0.00229

So, pH = -log([H+]) = -log(x) = 2.64

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