Respuesta :
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
The study of chemicals and bonds is called chemistry.
The correct answer is mentioned below.
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 be considered to be equal to the concentration of HBr and NaOH respectively.
What is titration?
- Titration is a common laboratory method of quantitative chemical analysis to determine the concentration of an identified analyte. A reagent, termed the titrant or titrator, is prepared as a standard solution of known concentration and volume.
The formula used to solve the question is as follows:-
Number of moles [tex](mol) = Concentration\ of\ solution (mol/L) * volume\ of\ solution (L)[/tex]
The moles of the following solution is as follows:-
moles of H+[tex]= 0.200 mol/L * 0.02 L = 0.004 moles[/tex]
moles of OH- = [tex]0.200 mol/L * 0.015 L = 0.003 mol[/tex]
moles of OH- [tex]= 0.200 mol/L * 0.0199 L = 0.00398 mol[/tex]
moles of OH- =[tex]0.200 mol/L * 0.020 L = 0.004 mol[/tex]
moles of OH-[tex]= 0.200 mol/L * 0.0201 L = 0.00402 mol[/tex]
moles of OH-[tex]= 0.200 mol/L * 0.035 L = 0.007 mol[/tex]
Before the neutralization point, moles of H+ have to be determined by taking the difference between moles of H+ in the initial solution and total moles of OH- added. After the neutralization point, moles of OH- have to be determined by taking the difference between total moles of OH- added and moles of H+ in the initial solution.
pH at neutralization point is 7Moles of H+/OH- remaining is as follows moles of
[tex]H+ = 0.004 - 0.003 = 0.001 mole[/tex]
[tex]moles of H+ = 0.004 - 0.00398 = 0.00002 mole[/tex]
[tex]moles of H+ = moles of OH- = 0.004 mole[/tex]
(neutralization point) moles of OH[tex]- = 0.00402 - 0.004 = 0.00002 mole\\moles\ of \ OH- = 0.007 - 0.004 = 0.003 mole[/tex]
The total volume of the solution has to be determined by adding the volume of the initial solution and the volume of titrating solution added. The concentration of H+/OH- has to be calculated by dividing moles of H+/OH- by the total volume on a solution.
Calculate the concentration of H+/OH- after addition of base
Total volume =[tex]0.002 + 0.0015 = 0.0035 L[/tex]
[H+][tex]= 0.001 mol / 0.0035 L = 0.286 mol/Lb)[/tex]
total volume[tex]= 0.002 + 0.00199 = 0.00399 L[/tex]
[H+] [tex]= 0.00002 mol / 0.00399 L = 0.00501 mol/Lc)[/tex]
Total volume[tex]= 0.002 + 0.002 = 0.004 L[H+] = [OH-] = 0.004 mol / 0.004 L = 1 mol/Ld)[/tex]
Total volume [tex]= 0.002 + 0.00201 = 0.00401 L[OH-] = 0.00002 mol / 0.00401 L = 0.00499 mol/Le)[/tex]
Total volume[tex]= 0.002 + 0.0035 = 0.0055 L[OH-] = 0.003 mol / 0.0055 L = 0.545 mol/L[/tex]
The formula to calculate pH from the concentration of H+ and OH- is:
[tex][pH = -log[H^+]pH = 14 - pOH = 14 + log[OH^-][/tex]
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) = 11.698e) pH = 14 + log(0.545) = 13.736
Hence, the correct answer is mentioned above.
For more information about titration, refer to the link:-https://brainly.com/question/2728613
