Question

Calculate the pH at the equivalence point in titrating 0.045 M solutions of each of the following with 0.061 M NaOH.

(a) hydrochloric acid (HCl) pH =

(b) acetic acid (HC2H3O2), Ka = 1.8e-05 pH =

(c) hypoiodous acid (HIO), Ka = 2.3e-11 pH =

Answer 1

A) When 0.045M HCl solution is titrated with 0.061M NaOH solution at equivalence point pH will be 7.

This is because HCl is strong acid and NaOH is a strong base at equivalence point it only forms water and salt.

HCl (aq) + NaOH (aq) H2O(l) + NaCl(aq)

pH=(1/2)pKw

pH=0.5*(-log(1*10-14)

pH=7(ans)

B) When Acetic acid (0.045M) is treated with NaOH (0.061M)

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O (l)

(weak acid)        (strong base)      (salt)               (water)

CH3COOH is a weak acid and NaOH is a strong base, for strong acid and weak base at equivalence point

now let the volume of CH3COOH is X ml and volume of NaOH is Y ml

so, we know V1S1=V2S2                 V1=x ml; V2= y ml ; S1= 0.045M; S2= 0.061M

X * 0.045 = Y * 0.061

at equivalence point total volume = X+0.738X=1.738X ml

Now 1000 ml solution Contains 0.045 mol CH3COOH

that is at equivalence point 1000ml Solution contains 0.045 mol CH3COONa

                                            1ml solution contains (0.045/1000) mol CH3COONa

                                            Xml Solution contains {(0.045/1000)*X} mol CH3COONa

now at equivalence concentration of CH3COONa is=( mol number of CH3COONa/ total volume)*1000              

                                                                         

                                                                       =0.026M

pKw= -log(1*10-14)=14

pKa= -log(1.8*10-5)=4.74

C= 0.026M

Now pH= (1/2)*(14+4.74+log(0.026))=8.57 (ans)

C) now for HOI,

HOI(aq) + NaOH(aq) NaOI(aq) + H2O(l)

HOI is weaker acid and NaOH is strong base, so

pKw= -log(1*10-14)=14

pKa= -log(2.3*10-11)=10.64

C= concentration of NaOI at equivalence point=0.026M ( using the similar method in question B)

now, pH= (1/2)(14+10.64+log(0.026))

       pH=11.52 (ans)