Question

A 98.0 mL sample of 0.0500 M HBr is titrated with 0.100 M CSOH solution. Calculate the pH after the following volumes of base have been added. (a) 14.2 mL (b) 47.5 mL (c) 49.0 mL pH = pH = pH (d) 51.0 mL (e) 80.4 mL pH = pH =

Answer 1

v

Given that

HBr = 0.0500 M

Volume = 98.0 ml

CsOH = 0.100 M

Reaction CsOH + HBr= CsBr + H2O

(A)

14.2 ml

Number of mole of HBr = molarity * volume in L

= 0.0500 *98/1000

= 0.0049 mole HBr

Mole of CsOH= 0.1*14.2/1000

0.00142 moles

Reming mole of HBr=0.0049 mole HBr-0.00142 moles

= 0.00348 moles HBr

Total volume = 98+14.2= 112.2 ml

Molarity = 0.00348 /0.1122

= 0.0310 M HBr or H+

pH = - log 0.0310= 1.50

(b)

47.5 ml

Number of mole of HBr = molarity * volume in L

= 0.0500 *98/1000

= 0.0049 mole HBr

Mole of CsOH= 0.1*47.5/1000

0.00475 moles

Reming mole of HBr=0.0049 mole HBr-0.00475 moles

= 0.00015 moles HBr

Total volume = 98+47.5= 145.5 ml

Molarity = 0.00015 /0.1455

= 1.03*10^-3 M HBr or H+

pH = - log 1.03*10^-3 = 2.98

(c)

49 ml

Number of mole of HBr = molarity * volume in L

= 0.0500 *98/1000

= 0.0049 mole HBr

Mole of CsOH= 0.1*49/1000

0.0049 moles

Reming mole of HBr=0.0049 mole HBr-0.0049 moles

= 0.0 moles HBr

Equivalence point p H = 7.00

(d)

51.0 ml

Number of mole of HBr = molarity * volume in L

= 0.0500 *98/1000

= 0.0049 mole HBr

Mole of CsOH= 0.1*51/1000

0.0051 moles

Reming mole of CsOH=0.0051 mole -0.0049 moles

= 0.0002 moles CsOH

Total volume = 98+51= 149 ml

Molarity of OH- = 0.0002 /0.149

= 1.34*10^-3 M CsOH or OH-

pOH = - log 1.34*10^-3 = 2.87

pH = 14- pOH

= 14- 2.87

= 11.13

(e)

80.4 ml

Number of mole of HBr = molarity * volume in L

= 0.0500 *98/1000

= 0.0049 mole HBr

Mole of CsOH= 0.1*80.4/1000

0.00804 moles

Reming mole of CsOH=0.00804 mole -0.0049 moles

= 0.00314 moles CsOH

Total volume = 98+80.4= 178.4 ml

Molarity of OH- = 0.00314 /0.1784

= 0.0176 M CsOH or OH-

pOH = - log 0.0176= 1.75

pH = 14- pOH

= 14- 1.75

= 12.25