Question

In a titration, you start with 0.05 M sodium hydroxide (NaOH) and slowly add it to 40 mL of HC1 with an unknown concentration. What is the original concentration of the HC1 if you add a total of 15.4 mL of sodium hydroxide to the HC1 to achieve pH =7?

Answer 1

The original concentration of the HCl is: 0.01925 M

Step-by-step explanation:

Equation:

HCl+NaOH-------> NaCl+H₂O

Volume of NaOH added = 0.05 M

No of moles of NaOH = 15.4 mL
`x(1 L)/(1000 mL) x(0.05 mol NaOH)/(L)`

= 0.00077 mol NaOH

Then,

Volume of HCl solution = 40 mL x 1/1000 mL

= 0.0400 L

Therefore,

Concentration of HCl = 0.00077 mol/0.0400 L

= 0.01925 M

Now, to find the pH:

pH = -log₁₀[H⁺]

= -log₁₀(2x10⁻⁶)

= 5.7