calculate the pH ofa solution prepared by mixing 50.0 ml of 0.200 m ch2h3o2 and 50 ml of 0.100 m Naoh.[ka(ch3cooh)=1.8*10^-5]
Yes
To calculate the pH of the solution, we need to determine the concentration of the common ion, which in this case is the acetate ion (CH3COO-).
First, let's calculate the moles of acetic acid (CH3COOH) and sodium hydroxide (NaOH) used in the solution.
Moles of CH3COOH = volume (in liters) × molarity = (50.0 ml / 1000 ml/L) × 0.200 mol/L
Moles of CH3COOH = 0.010 mol
Moles of NaOH = volume (in liters) × molarity = (50 ml / 1000 ml/L) × 0.100 mol/L
Moles of NaOH = 0.005 mol
Since NaOH is a strong base, it undergoes complete dissociation in water:
NaOH → Na+ + OH-
Therefore, after the reaction occurs, we will have the same number of moles of hydroxide ions (OH-) as we had moles of NaOH:
Moles of OH- = 0.005 mol
Now, let's analyze the reaction between acetic acid and hydroxide ions, which forms water and the acetate ion:
CH3COOH + OH- → CH3COO- + H2O
Since we had 0.010 moles of acetic acid and 0.005 moles of OH-, the reaction will consume all of the hydroxide ions and produce an equal amount of acetate ions:
Moles of CH3COO- = 0.005 mol
Next, we'll calculate the concentration of the acetate ion by dividing the moles by the total volume of the solution.
Total volume = volume of CH3COOH + volume of NaOH = 50.0 ml + 50.0 ml = 100.0 ml = 0.100 L
Concentration of CH3COO- = moles / total volume = 0.005 mol / 0.100 L
Concentration of CH3COO- = 0.050 M
Now, we can calculate the pOH of the solution using the concentration of hydroxide ions:
pOH = -log10[OH-]
pOH = -log10(0.050)
pOH ≈ 1.30
Since pH + pOH = 14, we can calculate the pH:
pH = 14 - pOH
pH = 14 - 1.30
pH ≈ 12.70
Therefore, the pH of the solution prepared by mixing 50.0 mL of 0.200 M CH3COOH and 50 mL of 0.100 M NaOH is approximately 12.70.
I don't know what you tried to write; my best guess is that you have CH3COOH and NaOH. I will call CH3COOH HAc.
millimols CH3COOH = 50 mL x 0.200 M = 10
millimols NaOH = 50 x 0.1 = 5
...........CH3COOH + NaOH ==> CH3COONa + H2O
I..............10................0.................0
add.............................5..............................
C..............-5..............-5...................+5
E...............5...............0.....................5
So you have a buffered solution. Use the Henderson-Hasselbalch equation.
Post your work if you get stuck.