A 0.080L solution of Ca(OH)2 is neutralized by 0.0293L of a 3.58 M H2CrO4 solution. What is the concentration of the Ca(OH)2 solution?

1.31 M

Explanation:

Step 1:

Data obtained from the question. This include the following:

Volume of Base (Vb) = 0.080L

Molarity of base (Mb) =..?

Volume of acid (Va) = 0.0293L

Molarity of acid (Ma) = 3.58 M

Step 2:

The balanced equation for the reaction. This is given below:

H2CrO4 + Ca(OH)2 → CaCrO4 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 3:

Determination of the concentration of base.

The concentration of the base can be obtained as follow:

MaVa/MbVb = nA/nB

3.58 x 0.0293 / Mb x 0.080 = 1

Cross multiply

Mb x 0.080 = 3. 58 x 0.0293

Divide both side by 0.080

Mb = (3.58 x 0.0293)/0.08

Mb = 1.31 M

Therefore, the concentration of the base, Ca(OH)2 is 1.31 M

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