1) How many grams of dry NH4Cl need to be added to 2.40 L of a 0.800 M solution
ID: 956272 • Letter: 1
Question
1) How many grams of dry NH4Cl need to be added to 2.40 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.82? Kb for ammonia is 1.8 x 10^-5.2) A beaker with 150 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.80 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
1) How many grams of dry NH4Cl need to be added to 2.40 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.82? Kb for ammonia is 1.8 x 10^-5.
2) A beaker with 150 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.80 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
1) How many grams of dry NH4Cl need to be added to 2.40 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.82? Kb for ammonia is 1.8 x 10^-5.
2) A beaker with 150 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.80 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
2) A beaker with 150 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.80 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
Explanation / Answer
1) we have formula pOH =pkb + log [NH4Cl]/[NH3]
where pkb = -log Kb = -log ( 1.8x10^-5) = 4.745 , pOH = 14-pH = 14-8.82 = 5.18
hence 5.18 = 4.745 + lg [NH4Cl] / ( 0.8)
[NH4Cl] = 2.17816
NH4Cl moles = M x V = 2.17816 x 2.4 = 5.2276
NH4Cl mass = moles x molar mass = 5.2276 x 53.491 = 279.63 g
2) pH = pka + log [acetate ]/[acetic acid]
5 = 4.74 + log [acetate ]/[acetic acid]
[acetate ] = 1.82 [ acetic acid] ...........(1)
given [acetate] + [acetic acid] = 0.1 ............(2)
by (1) ( 2) we get [acetic acid] =0.03546 , [acetate] = 0.064539
Moles of acetic acid = M x V = 0.03546 x 150/1000 = 0.005319
acetate moles = 0.064539 x 0.15 = 0.00968
HCl moles added= H+ moles = M x V = 0.49 x 6.8/1000 = 0.003332
H+ reacts with acetate and frms acetic acid
Hence acetic acid moles = 0.005319+0.003332 = 0.008651
acetate moles = 0.00968 - 0.003332 =0.006348
pH = 4.74 + log ( 0.0046348/0.008651)
= 4.47
pH change = 5-4.47 = 0.53
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