3. At the beginning of a titration to standardize the NaOH solution, Student A adjusted very carefully the initial burette volume to 0.00 mL, but he did not notice an important air bubble in the tip of the burette. At the end of the titration, the air bubble is gone. Explain the effect of that mistake on the calculated molarity MB. (Will the experimental MB calculated by Student A be higher or lower than the true MB value

The molarity of the solution will be higher.

Explanation:

Standarization is a process in which the molarity of the solution is made to be known. The concentration determined is exact by this process.

To standarize the solution of NaOH, first the acid used for titration is taken in a flask say 10 ml, indicator and water (in this case 50 ml). Then biuret is filled with NaOH of known molarity and initial volume is noted.

The air bubble not taken notice of in this case will lower the volume as molarity is inversely proportional to volume. When volume decreases molarity increases and vice-versa.

Formula used is

molarity =

The bubble will be formed with minimum volume thus volume is also lowered.

i think it is true since you can test the answer and can use observations to complete it.

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