A20.0 ml sample of 0.200 m hbr solution is titrated with 0.200 m naoh solution. calculate the ph of the solution after the following volumes of base have been added

a. 15.0 ml
b. 19.9 ml
c. 20.0 ml
d. 20.1 ml
e. 35.0 ml

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

the answer is the