Calculate the ph of a solution that is 0.30 M in ammonia (NH3) and 0.20 M in amm
ID: 551460 • Letter: C
Question
Calculate the ph of a solution that is 0.30 M in ammonia (NH3) and 0.20 M in ammonium chloride. Kb=1.76 x 10^-5Then calculate the ph if you add 10 mL of 1.0 M HCl to 50 mL of your solution in the problem above. Calculate the ph of a solution that is 0.30 M in ammonia (NH3) and 0.20 M in ammonium chloride. Kb=1.76 x 10^-5
Then calculate the ph if you add 10 mL of 1.0 M HCl to 50 mL of your solution in the problem above.
Then calculate the ph if you add 10 mL of 1.0 M HCl to 50 mL of your solution in the problem above.
Explanation / Answer
kb = 1.76*10^-5
Pkb = -logKb
= log1.76*10^-5
= 4.754
POH = Pkb + log[NH4Cl]/[NH3]
= 4.754 + log0.2/0.3
= 4.754-0.1760 = 4.578
PH = 14-POH
= 14-4.578 = 9.422 <<<<answer
no of moles of NH4Cl = molarity * volume in L
= 0.2*0.05 = 0.01 moles
no of moles of NH3 = molarity * volume in L
= 0.3*0.05 = 0.015 moles
no of moles of HCl = molarity *volume in L
= 1*0.01 = 0.01 moles
no of moles of NH4Cl after the addition of 0.01 moles of HCl = 0.01+0.01 =0.02 moles
no of moles of NH3 after the addition of 0.01 moles of HCl = 0.015-0.01 = 0.005 moles
POH = Pkb + log[NH4Cl]/[NH3]
= 4.754 + log0.02/0.005
= 4.754+0.602 = 5.356
PH = 14-POH
= 14-5.356 = 8.644 >>>answer
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