Question

A 35.00 mL sample of 0.2500 M HCl is titrated with 0.440 M NaOH. calculate the pH after the following amounts of base have been added

A.10.0 mL

B. 20.0 mL

C.30.00 mL

Answer 1

reaction :

HCl + NaOH --------> H2O + NaCl

first we will calculate the moles of Hcl35 mL sample of 0.2500 M HCl,

moles= molarity * Volume IN L,

Moles = 0.2500* 0.035 L (35 ml=0.035 L),

0.00875 moles of HCl,

Titration of this HCl with 10 ml of 0.440 M , NaOH,

1) HERE we will calculate the moles of NaOH in 0.440 M ,10 ml SOLUTION,

MOLES OF NaOH = MOLARITY * volume IN L,

= 0.440*0.010 L, (10 ML=0.010 L),

= 0.0044 moles of NaOH,

FOR one mole of HCl it is required to add one mole of NaOH (for neutralisation),

but, 0.00875 moles - 0.0044 moles = 0.00435 Moles HCl is Excess.,

MOLARITY = 0.00435/ 0.045 ( 35 ml +10 ml=45 ml total volume of solution =0.045 L),

MOLARITY = 0.0966 M HCl,

pH= -LOG H3O⁺= -LOG HCl,

pH= -log 0.0966 = 1.01 ( it is acidic),

2) for 20 ml NaOH (0.440 M),

WE KNOW THE MOLES OF HCl= 0.00875 MOLES,

here we can calculate the moles of 20 ML 0.440 M NaOH,

MOLES= 0.440* 0.020 L ( 020 ML=0.020 L),

= 0.0088 moles NaOH,

( 0.0088-0.00875= 0.00005 EXCESS NaOH ),

= 0.00005/0.055 L (TOTAL VOLUME IN L),

= 0.00090 M NaOH,

PH= -LOG OH^- = 3.04 ,

PH = POH- 3.04,

PH= 14- 3.04= 10.95 (BASIC)

3) 30.0 ML o0440 M NaOH , WE KNOW THE MOLES OF ACID = 0.00875 moles HCl,

here we will calculate the moles of NaOH,

MOLES= 0.440*0.030 L (30 ML=0.030 L),

=0.0132 MOLES OF NaOH,

(0.0132-0.00875)= 0.00445 EXCESS NaOH Moles,

0.00445/0.065 L, (Total volume 35 ml +30 ml=65 ML = 0.065 L),

= 0.06846 M NaOH,

pH= -log OH^- 0.06846,

= 1.16 ,

pH= pOH- 1.16,

= 14-1.16 = 12.84 (HIGHLY BASIC)