A solution formed by mixing 13.0 mL of 0.013 M Ba(OH)2 with 31.0 mL of 7.0 ✕ 10-3 M NaOH. Calculate [OH-]
mols Ba(OH)2 = M x L = ?
mols OH in Ba(OH)2 = 2x that = ?
mols Naoh = m x L = ?
final (OH^-) = total (OH^-)/total L
NaOH?
5. Calculate the pH of 1.0 x 10-10 M HCl
To calculate the [OH-] concentration in the mixed solution, we need to first determine the moles of OH- ions in the solution.
Step 1: Calculate the moles of Ba(OH)2:
Moles of Ba(OH)2 = Volume (in L) * Molarity
Moles of Ba(OH)2 = 13.0 mL * (1 L / 1000 mL) * 0.013 M
Moles of Ba(OH)2 = 0.00169 mol
Step 2: Calculate the moles of NaOH:
Moles of NaOH = Volume (in L) * Molarity
Moles of NaOH = 31.0 mL * (1 L / 1000 mL) * 7.0 * 10^-3 M
Moles of NaOH = 0.000217 mol
Step 3: Calculate the total moles of OH-:
Total moles of OH- = Moles of Ba(OH)2 + Moles of NaOH
Total moles of OH- = 0.00169 mol + 0.000217 mol
Total moles of OH- = 0.001907 mol
Step 4: Calculate the total volume of the mixed solution:
Total volume of the mixed solution = Volume of Ba(OH)2 + Volume of NaOH
Total volume of the mixed solution = 13.0 mL + 31.0 mL
Total volume of the mixed solution = 44.0 mL = 0.044 L
Step 5: Calculate the concentration of OH-:
[OH-] = Total moles of OH- / Total volume of the mixed solution
[OH-] = 0.001907 mol / 0.044 L
[OH-] = 0.0434 M
Therefore, the concentration of hydroxide ions ([OH-]) in the mixed solution is 0.0434 M.