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

Answer :

The specific heat of calorimeter is

The energy of combustion per mole of vanillin is

Explanation :

Part 1 :

First we have to calculate the energy released for 0.1625 g of benzoic acid.

Energy released = Energy released × Mass of benzoic acid

Energy released = (26.42 kJ/g) × (0.1625g)

Energy released = -4.293 kJ

Now we have to calculate the specific heat of calorimeter.

Heat released by the reaction = Heat absorbed by the calorimeter

where,

= heat released by the reaction = -4.293 kJ

= heat absorbed by the calorimeter

= specific heat of calorimeter = ?

= change in temperature =

Now put all the given values in the above formula, we get:

Thus, the specific heat of calorimeter is

Part 2 :

First we have to calculate the energy released by the reaction.

where,

= heat released by the reaction = ?

= heat absorbed by the calorimeter

= specific heat of calorimeter =

= change in temperature =

Now put all the given values in the above formula, we get:

Now we have to calculate the energy of combustion per mole of vanillin.

Molar mass of vanillin = 152.15 g/mole

Mass of vanillin = 0.2070 g

Thus, the energy of combustion per mole of vanillin is