Course= general chemistry2 Q1. 50.00 mL of 0.150 HCL is titrated with 20.00 mL o
ID: 903354 • Letter: C
Question
Course= general chemistry2Q1. 50.00 mL of 0.150 HCL is titrated with 20.00 mL of 0.375 M NaOH. Determine the pH of the solution. Q2. A solution contains 0.500 M HC2H3O2 and 0.400 M NaC2H3O2. Dteremine the pH of the solution after 0.020 mil of HCL are added to 250.0 mL of the solution. Q3. Determine the pH of a 0.146 M NH4Br solution. Q4. Calculate the pH of a 0.148 M Sr(OH)2 solution. Course= general chemistry2
Q1. 50.00 mL of 0.150 HCL is titrated with 20.00 mL of 0.375 M NaOH. Determine the pH of the solution. Q2. A solution contains 0.500 M HC2H3O2 and 0.400 M NaC2H3O2. Dteremine the pH of the solution after 0.020 mil of HCL are added to 250.0 mL of the solution. Q3. Determine the pH of a 0.146 M NH4Br solution. Q4. Calculate the pH of a 0.148 M Sr(OH)2 solution. Course= general chemistry2
Q2. A solution contains 0.500 M HC2H3O2 and 0.400 M NaC2H3O2. Dteremine the pH of the solution after 0.020 mil of HCL are added to 250.0 mL of the solution. Q3. Determine the pH of a 0.146 M NH4Br solution. Q4. Calculate the pH of a 0.148 M Sr(OH)2 solution.
Explanation / Answer
1) No of moles of HCl = 50/1000*0.15 = 0.0075 mole
No of moles of NaOH = 20/1000*0.375 = 0.0075 mole
as the No of moles of HCl, NaOH is same .the pH = 7
2) pH of acidic buffer = pka + log(salt-HCladded/acid+ HCl added)
pka of CH3COOH = 4.76
HCl added = 0.02 mol
= 4.76+log((0.4-0.02)/(0.5+0.02))
pH = 4.624
3) NH4Br is acidc buffer
pH = 7-1/2(pkb+logC)
pkb of NH4OH = 4.74
C = Concentration of salt = 0.146 M
= 7-1/2(4.74+log0.146)
= 5.045
4) Concentration of OH- = 2*0.148 = 0.296 M
pOH = -log0.296 = 0.53
pH = 14-0.53 = 13.47