In: Chemistry
A. A 75.0-mLmL volume of 0.200 MM NH3NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 MM HNO3HNO3. Calculate the pHpH after the addition of 11.0 mLmL of HNO3HNO3.
B. A 50.0-mLmL volume of 0.15 MM HBrHBr is titrated with 0.25 MM KOHKOH. Calculate the pHpH after the addition of 15.0 mLmL of KOHKOH.
C. A 52.0-mLmL volume of 0.35 MM CH3COOHCH3COOH (Ka=1.8×10−5Ka=1.8×10−5) is titrated with 0.40 MM NaOHNaOH. Calculate the pHpH after the addition of 17.0 mLmL of NaOHNaOH.
A. A 75.0-mLmL volume of 0.200 MM NH3NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 MM HNO3HNO3. Calculate the pHpH after the addition of 11.0 mLmL of HNO3HNO3.
pKb of NH3 = -log 1.8x10-5 = 4.75
NH3 + HNO3 -------------------> NH4NO3
75x0.2 =15 11x0.5= 5.5 0 initial mmoles
9.5 0 5.5 equilibrium
The pH of this buffer is calculated using Hendersen equation as
pOH = pKb + log [conjugate acid]/[base]
= 4.75 + log 5.5/9.5 = 4.513
pH = 14- pOH = 14-4.513 = 9.487
B. A 50.0-mLmL volume of 0.15 MM HBrHBr is titrated with 0.25 MM KOHKOH. Calculate the pHpH after the addition of 15.0 mLmL of KOHKOH.
BOth HBR and KOH are strong so no buffer formation in titration
HBr + KOH ---------------> KBr + H2O
50x0.15=7.5 15x0.25=3.75 0 0 initial mmoles
3.75 0 3.75 ----------- equilibrium
Solution has excess of acid and
[acid] = mmoles / total volume = 3.75 /(50+15) =0.0577 M
pH = -log [H+] = -log 0.0577 =1.24
C. A 52.0-mLmL volume of 0.35 MM CH3COOHCH3COOH (Ka=1.8×10−5Ka=1.8×10−5) is titrated with 0.40 MM NaOHNaOH. Calculate the pHpH after the addition of 17.0 mLmL of NaOHNaOH.
pKa of acetic acid = -log 1.8x10-5 = 4.75
CH3COOH + NaOH ------------------> CH3COONa + H2O
52x0.35=18.2 17x0.4=6.8 0 0 initial mmoles
11.4 0 6.8 -- equilibrium
The pH of this buffer is calculated using Hendersen equation as
pH = pKa + log [conjugate base ]/[acid]
= 4.75 + log 6.8/11.4
=4.525