Question

Find the pH of the following solutions given: pH = pKa + log (base/acid)

1) The titration of 450.00 mL of 0.5 M nitric acid with 0.3 M sodium hydroxide after 250.0 mL of sodium hydroxide has been added...

2)Given the K_b for aniline = 4.3 x 10^-3 at 25 degree Celsius, what is the pH of a .40 Molar solution of anilinium nitratate?

Answer 1

1)

no.of moles nitric acid = molarity*volume in litre = 0.5*0.45 = 0.225 moles

no.of moles of NaOH added is = 0.3*0.25lt = 0.075 moles

HNO3 + NaOH ............> NaNO3 + H2O

According to above chemical reaction

1 mole of HNO3 reacts with 1 mole of NaOH

0.075 moles of NaOH reacts with 0.075 moles of HNO3(0.225-0.075 = 0.15)

remaining no.of moles of HNO3 = 0.15 moles

Volume of the solution is = 450 ml + 250 ml = 700 ml

Molarity of HNO3 = no.of moles/volume in lt. = 0.15/0.7 = 0.214M

[H+] = [HNO3] = 0.214 M

Ph = -log[H+]

Ph = -log[0.214]

Ph = 0.67

2)

we know that

Ka*Kb = 10^-14

given Kb = 4.3 x 10^-3 and C = 0.40 M

Ka = 10^-14/4.3*10^-3

Ka = 2.32*10^-12

[H+] = [Ka*C]^1/2

       = [2.32*10^-12*0.4]^1/2

       = 9.6*10^-7 M

Ph = -log[H+]

Ph = -log[9.6*10^-7]

Ph = 6.02