Calculate [OH-] and pH for the following strong base solution;
a solution formed by mixing 10.0 mL of 0.011 M Ba(OH)2 with 28.0 mL of 7.4 10-3 M NaOH
working and answer would be appreciated
mol Ba(OH)2 = M x L = ??.
(OH^-) from Ba(OH)2 is twice that.
Add to the following.
mol NaOH = M x L = ?
(OH^-) = total mole OH^-/total L
Then pOH = -log(OH^-)
Substitute pOH in the following and solve for pH.
pH + pOH = pKw = 14
To calculate the concentration of the hydroxide ion ([OH-]) and the pH of the solution, we need to determine the total volume of the solution and the number of moles of hydroxide ions.
Step 1: Calculate the total volume of the solution.
The total volume of the solution is the sum of the volumes of Ba(OH)2 and NaOH solutions.
Total Volume = Volume of Ba(OH)2 + Volume of NaOH
Total Volume = 10.0 mL + 28.0 mL
Total Volume = 38.0 mL
Step 2: Determine the number of moles of hydroxide ions.
We need to consider the dissociation of both Ba(OH)2 and NaOH to calculate the number of moles of hydroxide ions.
Ba(OH)2(s) → Ba2+(aq) + 2OH-(aq) (1:2 mole ratio)
NaOH(aq) → Na+(aq) + OH-(aq) (1:1 mole ratio)
For Ba(OH)2:
Number of moles of OH- from Ba(OH)2 = Molarity of Ba(OH)2 x Volume of Ba(OH)2
Number of moles of OH- from Ba(OH)2 = 0.011 M x (10.0 mL / 1000 mL/L)
Number of moles of OH- from Ba(OH)2 = 0.00011 moles
For NaOH:
Number of moles of OH- from NaOH = Molarity of NaOH x Volume of NaOH
Number of moles of OH- from NaOH = 7.4 x 10^-3 M x (28.0 mL / 1000 mL/L)
Number of moles of OH- from NaOH = 0.0002072 moles
Total number of moles of OH- in the solution = Number of moles of OH- from Ba(OH)2 + Number of moles of OH- from NaOH
Total number of moles of OH- in the solution = 0.00011 moles + 0.0002072 moles
Total number of moles of OH- in the solution = 0.0003172 moles
Step 3: Calculate the concentration of [OH-].
Concentration of OH- = Total number of moles of OH- / Total volume of the solution
Concentration of OH- = 0.0003172 moles / (38.0 mL / 1000 mL/L)
Concentration of OH- = 0.0083505 M
Step 4: Calculate the pH of the solution.
To calculate the pH, we can use the formula:
pOH = -log10[OH-]
pH = 14 - pOH
pOH = -log10(0.0083505)
pOH ≈ 2.08
pH = 14 - 2.08
pH ≈ 11.92
Therefore, the [OH-] concentration is approximately 0.0083505 M, and the pH of the solution is approximately 11.92.
To calculate [OH-] and pH for the given strong base solution, we need to first determine the concentration of hydroxide ions ([OH-]) in the final solution.
Step 1: Calculate the number of moles of Ba(OH)2 in 10.0 mL of 0.011 M Ba(OH)2 solution.
Moles of Ba(OH)2 = Volume (in L) × Molarity
Moles of Ba(OH)2 = 0.010 L × 0.011 mol/L
Step 2: Calculate the number of moles of NaOH in 28.0 mL of 7.4 × 10-3 M NaOH solution.
Moles of NaOH = Volume (in L) × Molarity
Moles of NaOH = 0.028 L × 7.4 × 10-3 mol/L
Step 3: Calculate the total moles of hydroxide ions ([OH-]) in the final solution by adding the moles of Ba(OH)2 and NaOH.
Total moles of OH- = moles of Ba(OH)2 + moles of NaOH
Step 4: Calculate the total volume of the final solution by adding the volumes of Ba(OH)2 and NaOH.
Total volume of final solution = 10.0 mL + 28.0 mL
Step 5: Calculate the concentration of hydroxide ions ([OH-]) in the final solution.
[OH-] = Total moles of OH- / Total volume of solution
Step 6: Calculate the pOH using the concentration of hydroxide ions ([OH-]).
pOH = -log[OH-]
Step 7: Calculate the pH using the pOH.
pH = 14 - pOH
Now, let's calculate the results using the given values:
Step 1: Moles of Ba(OH)2 = 0.010 L × 0.011 mol/L = 0.00011 mol
Step 2: Moles of NaOH = 0.028 L × 7.4 × 10-3 mol/L = 2.072 × 10-4 mol
Step 3: Total moles of OH- = 0.00011 mol + 2.072 × 10-4 mol = 0.0002812 mol
Step 4: Total volume of final solution = 10.0 mL + 28.0 mL = 38.0 mL = 0.038 L
Step 5: [OH-] = 0.0002812 mol / 0.038 L = 0.00737 M
Step 6: pOH = -log(0.00737) = 2.13
Step 7: pH = 14 - 2.13 = 11.87
Therefore, the concentration of [OH-] in the final solution is 0.00737 M, and the pH is 11.87.