Calculate the pH of a 1.0 L solution that contains 0.50 M NH_3 and 0.40 M NH_4Cl
ID: 1021981 • Letter: C
Question
Calculate the pH of a 1.0 L solution that contains 0.50 M NH_3 and 0.40 M NH_4Cl. The K_b of NH_3 is 1.8 times 10^-5. A 1.0-L solution contains 0.50 M NH_3 and 0.40 M NH_4Cl. Calculate the pH of the solution after the addition of 100.0 mL of 1.0 M HCl. The K_b of NH_3 is 1.8 times 10^-5. 25.0 mL of 0.10 M formic acid (HCHO_2) is titrated with 0.10 M NaOH solution. Calculate the pH of the acid solution before any base has been added. The K_a of formic acid is 1.7 times 10^-4. 25.0 mL of 0.10 M formic acid (HCHO_2) is titrated with 0.10 M NaOH solution. Calculate the pH of the acid solution at the equivalence point of the titration. The k_a of formic acid is 1.7 times 10^-4.Explanation / Answer
32. POH = PKb + log[NH4Cl]/[NH3]
no of moles of NH4Cl = molarity * volume in L
= 0.4*1 = 0.4 moles
no of moles of NH3 = molarity * volume in L
= 0.5*1 = 0.5moles
Pkb = -logKb
= -log1.8*10-5
= 4.75
POH = Pkb + log[NH4Cl]/[NH3]
= 4.75 + log0.4/0.5
= 4.75-0.0969
= 4.6531
PH = 14-POH
= 14-4.6531
= 9.3469 >>>>> answer
33. By the addition of 100ml of 1 M HCl
no of moles of HCl = 1*0.1 = 0.1 moles
no of moles of NH4Cl = 0.4+0.1 = 0.5 moles
no of moles of NH3 = 0.5-0.1 = 0.4 moles
POH = Pkb + log[NH4Cl]/[NH3]
= 4.75 + log0.5/0.4
= 4.75 + 0.09691
= 4.8469
PH = 14-POH
= 14-4.8469 = 9.1531 >>>> answer
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