Question

What is the molarity of an HCl solution if 25.0 mL of this solution required 17.80 mL of 0.108 M NaOH to reach the end point in a titration?

Answer 1

`\boxed {\boxed {\sf 0.0769 \ M}}`

Step-by-step explanation:

We are asked to find the molarity of an acid given the details of a titration experiment. The formula for titration is as follows:

`M_AV_A= M_B V_B`

In this formula, M is the molarity of the acid or base and V is the volume of the acid or base. The molarity of the hydrochloric acid (HCl) is unknown and the volume is 25.0 milliliters.

`M_A * 25.0 \ mL = M_BV_B`

The molarity of the sodium hydroxide (NaOH) is 0.108 molar and the volume is 17.80 milliliters.

`M_A * 25.0 \ mL = 0.108 \ M * 17.80 \ mL`

We are solving for the molarity of the acid and we must isolate the variable
`M_A`. It is being multiplied by 25.0 milliliters. The inverse operation of multiplication is division, so we divide both sides of the equation by 25.0 mL.

`\frac {M_A * 25.0 \ mL }{25.0 \ mL}= (0.108 \ M * 17.80 \ mL )/(25.0 \ mL)`

`M_A= (0.108 \ M * 17.80 \ mL )/(25.0 \ mL)`

The units of milliliters cancel.

`M_A= (0.108 \ M * 17.80 )/(25.0 )`

`M_A= (1.9224)/(25.0 ) \ M`

`M_A= 0.076896 \ M`

The original measurements have 3 and 4 significant figures. We must round our answer to the least number of sig figs, which is 3. For the number we calculated, that is the ten-thousandth place. The 9 to the right of this place tells us to round the 8 up to a 9.

`M_A \approx 0.0769 \ M`

The molarity of the hydrochloric acid is 0.0769 Molar.