How much of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH
solution to the endpoint?
Show all work using the titrations formula.

2.5 ml of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH

solution to the endpoint

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

where,

are the n-factor, molarity and volume of acid which is

are the n-factor, molarity and volume of base which is KOH.

We are given:

Putting values in above equation, we get:

Thus 2.5 ml of 0.5 M HNO3 is necessary to titrate 25.0 mL of 0.05 M KOH

solution to the endpoint

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