Question

A 81.0 mL sample of 0.0300 M HIO4 is titrated with 0.0600 M RbOH solution. Calculate the pH after the following volumes of base have been added.

(a) 16.2 mL

pH =

(b) 38.9 mL

pH =

(c) 40.5 mL

pH =

(d) 40.9 mL

pH =

(e) 68.0 mL

pH =

Please Explain!!! I really need help

Answer 1

1)when 16.2 mL of RbOH is added

Given:

M(HIO4) = 0.03 M

V(HIO4) = 81 mL

M(RbOH) = 0.06 M

V(RbOH) = 16.2 mL

mol(HIO4) = M(HIO4) * V(HIO4)

mol(HIO4) = 0.03 M * 81 mL = 2.43 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.06 M * 16.2 mL = 0.972 mmol

We have:

mol(HIO4) = 2.43 mmol

mol(RbOH) = 0.972 mmol

0.972 mmol of both will react

remaining mol of HIO4 = 1.458 mmol

Total volume = 97.2 mL

[H+]= mol of acid remaining / volume

[H+] = 1.458 mmol/97.2 mL

= 1.5*10^-2 M

use:

pH = -log [H+]

= -log (1.5*10^-2)

= 1.8239

Answer: 1.82

2)when 38.9 mL of RbOH is added

Given:

M(HIO4) = 0.03 M

V(HIO4) = 81 mL

M(RbOH) = 0.06 M

V(RbOH) = 38.9 mL

mol(HIO4) = M(HIO4) * V(HIO4)

mol(HIO4) = 0.03 M * 81 mL = 2.43 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.06 M * 38.9 mL = 2.334 mmol

We have:

mol(HIO4) = 2.43 mmol

mol(RbOH) = 2.334 mmol

2.334 mmol of both will react

remaining mol of HIO4 = 9.6*10^-2 mmol

Total volume = 119.9 mL

[H+]= mol of acid remaining / volume

[H+] = 9.6*10^-2 mmol/119.9 mL

= 8.007*10^-4 M

use:

pH = -log [H+]

= -log (8.007*10^-4)

= 3.0965

Answer: 3.10

3)when 40.5 mL of RbOH is added

Given:

M(HIO4) = 0.03 M

V(HIO4) = 81 mL

M(RbOH) = 0.06 M

V(RbOH) = 40.5 mL

mol(HIO4) = M(HIO4) * V(HIO4)

mol(HIO4) = 0.03 M * 81 mL = 2.43 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.06 M * 40.5 mL = 2.43 mmol

We have:

mol(HIO4) = 2.43 mmol

mol(RbOH) = 2.43 mmol

2.43 mmol of both will react to form neutral solution

hence pH of solution will be 7

Answer: 7.00

4)when 40.9 mL of RbOH is added

Given:

M(HIO4) = 0.03 M

V(HIO4) = 81 mL

M(RbOH) = 0.06 M

V(RbOH) = 40.9 mL

mol(HIO4) = M(HIO4) * V(HIO4)

mol(HIO4) = 0.03 M * 81 mL = 2.43 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.06 M * 40.9 mL = 2.454 mmol

We have:

mol(HIO4) = 2.43 mmol

mol(RbOH) = 2.454 mmol

2.43 mmol of both will react

remaining mol of RbOH = 2.4*10^-2 mmol

Total volume = 121.9 mL

[OH-]= mol of base remaining / volume

[OH-] = 2.4*10^-2 mmol/121.9 mL

= 1.969*10^-4 M

use:

pOH = -log [OH-]

= -log (1.969*10^-4)

= 3.7058

use:

PH = 14 - pOH

= 14 - 3.7058

= 10.2942

Answer: 10.29

5)when 68.0 mL of RbOH is added

Given:

M(HIO4) = 0.03 M

V(HIO4) = 81 mL

M(RbOH) = 0.06 M

V(RbOH) = 68 mL

mol(HIO4) = M(HIO4) * V(HIO4)

mol(HIO4) = 0.03 M * 81 mL = 2.43 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.06 M * 68 mL = 4.08 mmol

We have:

mol(HIO4) = 2.43 mmol

mol(RbOH) = 4.08 mmol

2.43 mmol of both will react

remaining mol of RbOH = 1.65 mmol

Total volume = 149.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 1.65 mmol/149.0 mL

= 1.107*10^-2 M

use:

pOH = -log [OH-]

= -log (1.107*10^-2)

= 1.9557

use:

PH = 14 - pOH

= 14 - 1.9557

= 12.0443

Answer: 12.04