Question

A 25.00 ml sample of hydrochloric acid solution, hcl, is titrated with 0.0512 m naoh solution. the volume of naoh solution required is 21.68 ml. what is the molarity of the hcl solution?

Answer 1

A 25.00 ml sample of hydrochloric acid solution, HCl, is titrated with 0.0512 m NaOH solution. the volume of NaOH solution required is 21.68 ml  then the molarity of the HCl solution is 0.044 M .

Calculation ,

Formula used : $M_{1} V_{1} =M_{2} V_{2}$                            ...( i )

Where M is the molarity or concentration and V is the volume in ml .

concentration of hydrochloric acid solution ( $M_{1}$    ) = ?

concentration of NaOH ( $M_{2}$ ) = 0.0512 M

volume of hydrochloric acid solution  ( $V_{1}$ ) = 25.00 ml

volume of NaOH ( $V_{2}$ ) =  21.68 ml

Putting the value of concentration , volume of both  in equation ( i ) we get .

$M_{1}$   × 25.00 ml = 0.0512 × 21.68 ml

$M_{1}$    = 0.0512 × 21.68 ml / 25.00 ml= 0.044 M

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