Question

5 ) A 15.00 mL sample of an unknown monoprotic weak acid solution is titrated with 0.35 M NaOH . The initial buret reading is 0.23 mL . The phenolphthalein indicator turns the solution light pink when the buret reads of 29.58 mL . What volume of 0.35 M NaOH was delivered ? b . How many moles of NaOH were delivered ? How many moles of weak acid were present in the solution ? d . Calculate the molarity of the unknown weak acid solution ? a . C.

Answer 1

(a) 29.35 mL

(b) 0.0102725 moles

(c) 0.010272 moles

(d) 0.6848 M

Step-by-step explanation

Given:

Volume of the unknown acid, Va = 15.00 mL

Molarity of NaOH solution, Cb = 0.35 M

Initial buret reading = 0.23 mL

Final buret reading = 29.58 mL

Step-by-step solution:

(a) What volume of 0.35 M NaOH was delivered?

Volume of 0.35 M NaOH solution used, Vb = (Final buret reading - Initial buret reading) = 29.58 mL - 0.23 mL = 29.35 mL

(b) How many moles of NaOH were delivered?

The number of moles of NaOH delivered can be calculated using the molarity formula.

`Molarity=\frac{Mole}{Volume\text{ }in\text{ }L}`

Putting molarity of NaOH = 0.35 M and volume of NaOH = 29.35/1000 = 0.02935 L into the formula, we have;

`Mole=Molarity* Volume=0.35M*0.02935L=0.0102725\text{ }mol`

(d) Calculate the molarity of the unknown weak acid solution.

Since the unknown acid is a monoprotic weak acid, therefore, the mole ratio of the acid to the base is 1:1.

Hence, the molarity of the unknown weak acid solution can be calculated using the formula below

`\begin{gathered} (C_aV_a)/(n_a)=(C_bV_b)/(n_b) \\ \\ \Rightarrow C_a=(C_bV_bn_a)/(V_an_b) \end{gathered}`

Ca is the molarity of the unknown acid, put Va = 15.0 mL, Vb = 29.35 mL, Cb = 0.35 M, na = 1 and nb = 1 into the formula.

`C_a=(0.35M*29.35mL*1)/(15.0mL*1)=(10.2725M)/(15.0)=0.6848\text{ }M`

Therefore, the molarity of the unknown weak acid is 0.6848 M.

(c) How many moles of weak acid were present in the solution?

The number of moles of weak acid present in the solution can be calculated using the molarity formula in part (b) above.

Molarity of the weak acid = 0.6848 M and Volume in L = 15/1000 = 0.015 L

`\begin{gathered} Molarity=\frac{Mole}{Volume\text{ }in\text{ }L} \\ \\ \Rightarrow Mole=0.6848M*0.015L=0.010272\text{ }mol \end{gathered}`