Question

The reaction of HCl with NaOH is represented by the equation
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
What volume of 0.417 M HCl is required to titrate 11.9 mL of 0.151 M NaOH?
a. 4.3 mL
b. 2.39 mL
c. 0.749 mL
d. 32.8 mL
e. 11.9 mL

Answer 1

Answer:

a. 4.3 mL

Explanation:

Step 1: Write the balanced neutralization reaction

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Step 2: Calculate the reacting moles of NaOH

11.9 mL of 0.151 M NaOH react.

0.0119 L × 0.151 mol/L = 0.00180 mol

Step 3: Calculate the reacting moles of HCl

The molar ratio of HCl to NaOH is 1:1. The reacting moles of HCl are 1/1 × 0.00180 mol = 0.00180 mol.

Step 4: Calculate the volume of 0.417 M HCl that contains 0.00180 moles of HCl

0.00180 mol × (1000 mL/0.417 mol) = 4.3 mL

Answer 2

The volume of HCL will be "4.3 mL".

According to the question,

Mass,

• $M_1 = 0.417$
• $M_2 = 11.9$

Volume,

• $V_1 = ?$
• $V_2 = 0.151$

As we know the relation,

→ $M_1V_1 = M_2 V_2$

or,

→      $V_1 = \frac{M_2 V_2}{M_1}$

By substituting the values, we get

            $= \frac{11.9\times 0.151}{0.417}$

            $= \frac{1.7969}{0.417}$

            $= 4.3 \ mL$

Thus the above answer is right.

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