How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid? How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid?

a. 37.50 mL

b. 50.00 mL

c. 75.00 mL

d. 100.00 mL

e. 25.00 mL

We need 75 mL of 0.1 M NaOH (Option C)

Explanation:

Step 1: Data given

Molarity of NaOH solution = 0.100 M

volume of 0.150 M CH3COOH = 50.00 mL = 0.05 L

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles of CH3COOH

Moles CH3COOH = Molarity * volume

Moles CH3COOH = 0.150 M * 0.05 L

Moles CH3COOH = 0.0075 moles

Step 4: Calculate moles of NaOH

For 1 mol of CH3COOH we need 1 mol of NaOH

For 0.0075 mol CH3COOH we need 0.0075 mole of NaOH

Step 5: Calculate volume of NaOH

volume = moles / molarity

volume = 0.0075 moles / 0.100 M

Volume = 0.075 L = 75 mL

We need 75 mL of 0.1 M NaOH