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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...

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.

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Expert Solution

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

queen_honey_blossom answered 5 months ago

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