Question

Consider the titration of 100.0 mL of 0.100 M NaOH with 1.00 M HBr. Find the pH at the following volumes of acid added: Va = 0, 1.0, 2.5, 6.5, 9.9, 10.0, 10.1, and 11.5 mL.

Answer 1

NaOH and HBr are strong, and will dissociate completely in solition

NaOH --> Na+a and OH-

HBr --> H+ and Br-

M = 0.1 NaOH

V = 100 ml

M2 = 1 HBr

V2 = ?

Find initial moles of NaOH:

M1V1 = 100*0.1 = 10 mmol of NaOH

Now... find initial moles of HBr in the next quantities of volume:

M2*V2

V2 =0, 1.0, 2.5, 6.5, 9.9, 10.0, 10.1, and 11.5 mL.

this is eas since M = 1 M...

mmol HBr = 0, 1.0, 2.5, 6.5, 9.9, 10.0, 10.1, and 11.5 mmol of HBr

Underline --> means NaOH quantities is gretaer, therefore expec pH > 7

Bold --> means this is equivalence point, expec pH = 7

Italic --> means this is excess of acid, expect pH < 7

Now, let us calculate the total volumes of each experiment

VT = V1+V2 = 100 ml + V2

VT = 100, 101, 102.5, 106.5, 109.9, 110.0, 110.1, and 111.5 mL. (we will need this for concnetrations)

Now, calculate the final values of NaOH/HBr:

For NaOH in excess (underlined)

Mmol excess = 10 mmol of NaOH - X mmol of HBR

mmol NAOH = 10mmol - (0, 1.0, 2.5, 6.5, 9.9)

mmol NaOH = 10, 9, 7.5, 3.5, 0.1

There is no need to calculate the bold, since pH = 7 (neutralization)

for excess of HCl:

mmol of HCl: X - 10 = 10.1,11.5 - 10 = 0.1, 1.5 mmol of HCl

mmool of HCl = 0.1, 1.5

Now, calculate pH

From NaOh, is easier to calculate [NAOH] first, then pOH, then use pH = 14-pOH

Calculate concnetrations:

M = mmol/ml

mmol NaOH = 10, 9, 7.5, 3.5, 0.1

ml = 100, 101, 102.5, 106.5, 109.9,

Set of cocnetrations: 10/100, 9/101, 7.5/102.5, 3.5/106.5, 0.1/109.9

Set of [NaOH] = 0.1, 0.089, 0.0732, 0.0329, 0.00091

pOH = -log(OH-) = -log( 0.1, 0.089, 0.0732, 0.0329, 0.00091) = 1, 1.05, 1.14, 1.43, 3.04

pH = 14-pOH = 14, 12.95, 12.86, 12.52, 10.95

Now for the balance

pH = 7

Now for HBr excess:

calculate concnetration of HBr:

mmool of HCl = 0.1, 1.5

M = mmol/ml

M = (0.1, 1.5 )110.1, 11.5 = 9.1*10^-4 and 0.013

Calculate pH

pH = -log(H) = -log(9.1*10^-4, 0.013) = 3.04, 1.89

Therefore:

pH = 14, 12.95, 12.86, 12.52, 10.95, 7, 3.04, 1.89