85.0 mL of 0.25 M HF (Ka = 7.2 x 10-4)is titrated with 5.0 mL of 0.80 M NaOH. Calculate the pH at this point in the titration.
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well in order to answer this question you have to provide more information < 3

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

the answer to this question is 3.

it is the most commonly accepted one