Question

NaOH is used to titrate 15.2ml of HCl with unknown concentration. The indicator phenolphthalein is used to change color at the endpoint. The reading on the buret starts at 1.05ml and ends at 34.7ml. The concentration of NaOH had been determined to be 0.140M. Find the concentration of the acid. (explain work)

Answer 1

Q150860

NaOH is used to titrate 15.2ml of HCl with an unknown concentration. The indicator phenolphthalein is used to change color at the endpoint. The reading on the burette starts at 1.05ml and ends at 34.7ml. The concentration of NaOH had been determined to be 0.140M. Find the concentration of the acid. (explain work)

Solution:

NaOH is present in the burette. The initial burette reading is 1.05mL and the final burette reading is 34.7 mL.

So volume of NaOH used in titration = Final burette reading – Initial burette reading

= 34.7 mL – 1.05 mL

= 33.65 mL

The concentration of NaOH is already given to us, which is 0.140 M.

The volume of HCl used in the Titration = 15.2 mL.

The reaction of NaOH and HCl is given as

NaOH(aq) + HCl (aq) ----> NaCl (aq) + H₂O(l)

Step 1: Using the formula of molarity find the moles of NaOH.

$Molarity = \frac{moles \space of \space NaOH}{volume \space of \space solution \space in\space 'L' }$

33.65 mL in [ L' = $33.65 mL * \frac{1L}{1000mL} = 0.03365 L$

substitute 0.03365 L and 0.140 M in the formula we have

$0.140 mol/L = \frac{moles \space of \space NaOH}{0.03365L }$ ; (∵  M = mol/L)

multiply both the side by 0.03365 L, we have

0.140 mol/L * 0.03365L = moles of NaOH;

moles of NaOH = 0.004711moles.

Step 2: To find the moles of HCl that will react with 0.004711 moles of NaOH.

NaOH(aq) + HCl (aq) ----> NaCl (aq) + H₂O(l)

The mole to mole ratio of NaOH and HCl is 1 :1

Hence moles of HCl required = $0.004711 \space moles \space of \space NaOH * \frac{1\space mol \space HCl }{1\space mol \space NaOH}$ ;

= 0.004711 moles of HCl.

Step 3: To find the concentration of HCl.

We will use the formula of molarity again and find the concentration of HCl

$Molarity = \frac{moles \space of \space HCl }{volume \space of \space solution \space in\space 'L' }$

15.2 mL in 'L' = $15.2 mL * \frac{1L}{1000mL} = 0.0152 L;$

plug 0.0152 L and 0.004711 mol HCl in the formula we have

$Molarity = \frac{0.004711 \space mol\space HCl }{0.0152L};$

= 0.3099 mol / L or 0.3099 M;

In question, we are given all the quantities in 3 significant figure, so our final answer must also

be in 3 significant figures.

Hence the concentration of acid (HCl solution) is 0.310 M.

Method 2: You can also directly put the information in the dilution equation and find the concentration of HCl.

$M_{HCl}*V_{HCl} = M_{NaOH} * V_{NaOH};$

$M_{HCl}*15.2mL = 0.140M * 33.65mL ;$

divide both the side by 15.2 mL, we have

$M_{HCl} = \frac{ 0.140M * 33.65mL}{15.2mL};$

$M_{HCl} = 0.310 M;$