Use the following table of volumes and concentrations of strong acids and bases to determine the final pH for each solution (a - e). Use three significant figures in your answer.
Acid Base
Volume (L) Conc. [M] Volume (L) Conc. [M]
a) 1.760 0.384 0.834 0.724
b) 0.493 1.20 0.375 1.20
c) 0.038 0.0055 0.293 0.0010
d) 0.294 0.035 0.015 0.38
e) 5.013 5.7×10-4 0.395 3.4×10-1
It's difficult to interpret with improper spacing. Perhaps this is better. I'll do the first one and leave the others for you. Post your work if you get stuck. I'll call the strong acid HA and the strong base BOH.
.........Acid.........................................................................Base
Volume (L) .............Conc. [M] ..................Volume (L)............ Conc. [M]
a)... 1.760.................. 0.384....................... 0.834...................... 0.724
b) 0.493 1.20 0.375 1.20
c) 0.038 0.0055 0.293 0.0010
d) 0.294 0.035 0.015 0.38
e) 5.013 5.7×10-4 0.395 3.4×10-1
.....................HA + BOH ==> AB + H2O
mols HA = M x L = 1.760 x 0.384 = 0.676
mols BOH = M x L = 0.834 x 0.724 = 0.604
mols HA left in excess is 0.676 - 0.604 = 0.0720 mols in a volume of 1.760 L + 0.834 L = 2.59 L so concentration of the HA remaining is M = mols/L = 0.0720 mols/2.59 L = 0.0278 M. pH = -log(H^+) = -log(0.0278) = 1.56
Can I assume that the acid is monoprotic?
To determine the final pH for each solution, we need to calculate the resulting concentration of the strong acid or base in the solution after the reaction takes place. The reaction between a strong acid and a strong base results in the formation of water and a salt. Since we are assuming the volumes are added together, we can use the formula:
moles of acid = Volume (L) x Concentration (M)
Based on this, we can follow these steps to find the final pH for each solution:
a) Solution a:
Moles of acid = 1.760 L x 0.384 M = 0.67584 mol
Moles of base = 0.834 L x 0.724 M = 0.603216 mol
Since moles of acid > moles of base, the solution will be acidic. To find the final concentration of the acid, we subtract the moles of the base from the moles of the acid and divide by the total volume:
Final moles of acid = (0.67584 mol - 0.603216 mol) = 0.072624 mol
Final concentration of acid = 0.072624 mol / (1.760 L + 0.834 L) = 0.025 M
To find the pH, we use the formula:
pH = -log10(concentration of H+)
pH = -log10(0.025) ≈ 1.60
Therefore, the final pH of solution a is approximately 1.60.
b) Solution b:
Moles of acid = 0.493 L x 1.20 M = 0.5916 mol
Moles of base = 0.375 L x 1.20 M = 0.45 mol
Since moles of acid > moles of base, the solution will be acidic. Using the same calculations as in solution a:
Final moles of acid = (0.5916 mol - 0.45 mol) = 0.1416 mol
Final concentration of acid = 0.1416 mol / (0.493 L + 0.375 L) = 0.20 M
pH = -log10(0.20) ≈ 0.70
Therefore, the final pH of solution b is approximately 0.70.
c) Solution c:
Moles of acid = 0.038 L x 0.0055 M = 0.000209 mol
Moles of base = 0.293 L x 0.0010 M = 0.000293 mol
Since moles of base > moles of acid, the solution will be basic. Following the same calculations as before:
Final moles of base = (0.000293 mol - 0.000209 mol) = 0.000084 mol
Final concentration of base = 0.000084 mol / (0.038 L + 0.293 L) = 0.00028 M
pOH = -log10(0.00028) ≈ 3.55
pH = 14 - pOH ≈ 10.45
Therefore, the final pH of solution c is approximately 10.45.
d) Solution d:
Moles of acid = 0.294 L x 0.035 M = 0.01029 mol
Moles of base = 0.015 L x 0.38 M = 0.0057 mol
Since moles of acid > moles of base, the solution will be acidic. Using the same calculations as before:
Final moles of acid = (0.01029 mol - 0.0057 mol) = 0.00459 mol
Final concentration of acid = 0.00459 mol / (0.294 L + 0.015 L) = 0.015 M
pH = -log10(0.015) ≈ 1.82
Therefore, the final pH of solution d is approximately 1.82.
e) Solution e:
Moles of acid = (5.013 L x 5.7x10^-4 M) = 0.00285741 mol
Moles of base = (0.395 L x 3.4x10^-1 M) = 0.1343 mol
Since moles of base > moles of acid, the solution will be basic. Following the same calculations as before:
Final moles of base = (0.1343 mol - 0.00285741 mol) = 0.13144259 mol
Final concentration of base = 0.13144259 mol / (5.013 L + 0.395 L) = 0.025 M
pOH = -log10(0.025) ≈ 1.60
pH = 14 - pOH ≈ 12.40
Therefore, the final pH of solution e is approximately 12.40.
To determine the final pH for each solution, we need to use the concept of neutralization reaction between acids and bases. In a neutralization reaction, an acid reacts with a base to form water and a salt. The pH of the final solution depends on the concentrations of the acid and base used.
The pH of a solution can be calculated using the equation:
pH = -log[H+]
Where [H+] is the concentration of hydrogen ions in the solution.
Let's solve each scenario step by step:
a) In this case, you have an acid with a volume of 1.760 L and a concentration of 0.384 M, and a base with a volume of 0.834 L and a concentration of 0.724 M.
To determine the final concentration of hydrogen ions, we can use the formula:
(moles of H+ in acid) / (total volume of the solution)
moles of H+ in acid = volume of acid (L) * concentration of acid (M)
total volume of the solution = volume of acid + volume of base
moles of H+ in acid = 1.760 * 0.384
total volume of the solution = 1.760 + 0.834
Now we can calculate the concentration of hydrogen ions:
[H+] = (moles of H+ in acid) / (total volume of the solution)
Plug in the values and calculate [H+].
Finally, calculate the pH by taking the negative logarithm of [H+] using the pH equation:
pH = -log[H+]
Repeat these steps for scenarios b) to e) using the given values for volumes and concentrations of acids and bases.
Note: Make sure to use correct significant figures (three significant figures in this case) to maintain the appropriate level of precision in your final answer.