Answer:
a) pH = 0.544
b) pH = 2.300
c) pH = 7
d) pH = 11.698
e) pH = 13.736
Explanation:
Both HBr and NaOH are strong acids and bases so they can be considered to be fully dissociated in solution. Therefore the concentration of H+ and OH- can considered to be equal to the concentration of HBr and NaOH respectively.
Number of moles (mol) = Concentration of solution (mol/L) * volume of solution (L)
Step 1: Calculate number of moles of H+ in initial solution
moles of H+ = 0.200 mol/L * 0.02 L = 0.004 mol
Step 2: Calculate number of moles of OH- in titrating solution
a) moles of OH- = 0.200 mol/L * 0.015 L = 0.003 mol
b) moles of OH- = 0.200 mol/L * 0.0199 L = 0.00398 mol
c) moles of OH- = 0.200 mol/L * 0.020 L = 0.004 mol
d) moles of OH- = 0.200 mol/L * 0.0201 L = 0.00402 mol
e) moles of OH- = 0.200 mol/L * 0.035 L = 0.007 mol
Before neutralization point, moles of H+ have to be determined by taking the difference between moles of H+ in initial solution and total moles of OH- added. After neutralization point, moles of OH- have to be determined by taking the difference between total moles of OH- added and moles of H+ in initial solution. pH at neutralization point is 7
Step 3: Calculate moles of H+/OH- remaining
a) moles of H+ = 0.004 - 0.003 = 0.001 mol
b) moles of H+ = 0.004 - 0.00398 = 0.00002 mol
c) moles of H+ = moles of OH- = 0.004 mol (neutralization point)
d) moles of OH- = 0.00402 - 0.004 = 0.00002 mol
e) moles of OH- = 0.007 - 0.004 = 0.003 mol
Total volume of solution has to be determined by adding volume of initial solution and volume of titrating solution added. Concentration of H+/OH- has to be calculated dividing moles of H+/OH- by the total volume on solution.
Step 4: Calculate concentration of H+/OH- after addition of base
a) total volume = 0.002 + 0.0015 = 0.0035 L
[H+] = 0.001 mol / 0.0035 L = 0.286 mol/L
b) total volume = 0.002 + 0.00199 = 0.00399 L
[H+] = 0.00002 mol / 0.00399 L = 0.00501 mol/L
c) total volume = 0.002 + 0.002 = 0.004 L
[H+] = [OH-] = 0.004 mol / 0.004 L = 1 mol/L
d) total volume = 0.002 + 0.00201 = 0.00401 L
[OH-] = 0.00002 mol / 0.00401 L = 0.00499 mol/L
e) total volume = 0.002 + 0.0035 = 0.0055 L
[OH-] = 0.003 mol / 0.0055 L = 0.545 mol/L
The formula to calculate pH from concentration of H+ and OH- is:
pH = -log[H+]
pH = 14 - pOH = 14 + log[OH-]
Step 5: Calculate pH
a) pH = -log(0.286) = 0.544
b) pH = -log(0.00501) = 2.300
c) pH = 7 (neutralization point)
d) pH = 14 + log(0.00499) = 11.698
e) pH = 14 + log(0.545) = 13.736
