Question

1) Suppose 500 mL of an 0.200 molar HCl acid is titrated with 70.0 mL of a 1.00 M NaOH solution. What is the final pH of the solution? The error interval is 0.1 pH units.

2) Suppose 500 mL of 0.100 M benzoic acid, which has a pK_a of 4.201, solution is titrated with 75.0 mL of a 1.00 M NaOH. What is the final pH? The error interval is +/- 0.2 pH units.

Answer 1

Answer : Here 1} the required equation is like

HCl + NaOH <----------------------> NaCL + H2O

Number of moles 500 ml of Hcl = 0.200 * 500 / 1000 = 0.1 mol

Number of moles of 70 ml of NaoH = 1.00 * 70 / 1000 = 0.07 mol

Hence it is a acid base reaction hence the concentration of salt form is 0.07 mol

Now using Henderson's equations we can calculate the required relation

Ph = PKa + Log [salt] /[acid]

[NOTE : Here we have required the value of Ka in question ... ]

Now Put all the values and calculate the Ph .... so this is the basic procedure .

2} Here we have given the value of PKa .

Note : do the same procedure as above

Key Point find the number of mole

number of moles of 500 ml of benzoic acid = 0.100 * 500 /1000 = 0.05 mol

now , number of moles of 75 ml of NaOh = 1.00 * 75 / 1000 = 0.075 mol

Here again it is a acid base reaction means equal number acid can neutralyse the equal number of base and form the salt

Hence concentration of salt is = 0.05 mol

Now using Henderson's equation we get

Ph = 4.201 + log [0.05]/[0.05] = 4.201

Hence, this all about the question .