Chapter 18, Problem 81QAP

Interpretation Introduction

Interpretation:

The number of years should be calculated which takes a 1.00 M solution of U to reach pH of 6.0.

Concept introduction:

Activity is defined as the product of rate constant and number of atoms. It is also defined as the rate of decay or expresses number of atoms decays in unit time.

The mathematical expression is given by:

$\text{A = kN}$

Where, A = activity, k = first order rate constant and N = number of radioactive nuclei.

Activity can be expressed in following units:

1 Bq = 1 atom/s

1 Ci = 3.700 × 10¹⁰ atoms/s

To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of ${\text{[H}}^{+}\text{]}$ in aqueous solution is small, by using the p scale in the form of $\text{pH}$ scale, it is better way to represent acidity of solution.

Thus, $\text{pH}$ = ${\text{-log [H}}^{+}\text{]}$

The relation between half-life and decay constant is given by:

$\lambda \text{=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}$

Where, $\lambda$ = decay constant and ${\text{t}}_{\text{1/2}}$ = half-life

Answer to Problem 81QAP

The 1.00 M solution of U will take $t=3.03\times {10}^6\text{ year}$ to achieve pH = 6.0.

Explanation of Solution

Given information:

Half-life of uranium = $7\times {10}^8\text{ years}$

The given equation is:

${\text{Zn(H}}_{\text{2}}{\text{O)}}_{\text{4}}^{\text{2+}}\text{(aq)}\rightleftharpoons {\text{H}}^{\text{+}}{\text{(aq)+Zn(H}}_{\text{2}}{\text{O)}}_{\text{3}}{\text{(OH)}}^{\text{+}}{\text{(aq) K}}_a=3.3\times {10}^{-10}$

The reaction between zinc ions and water is given as:

${\text{Zn(H}}_{\text{2}}{\text{O)}}_{\text{4}}^{\text{2+}}\text{(aq)}\rightleftharpoons {\text{H}}^{\text{+}}{\text{(aq)+Zn(H}}_{\text{2}}{\text{O)}}_{\text{3}}{\text{(OH)}}^{\text{+}}\text{(aq) }$

Given pH value is 6.0.

$\text{pH}$ = ${\text{-log [H}}^{+}\text{]}$

Put the value of pH in above formula.

$\text{6}{\text{.0 = -log [H}}^{+}\text{]}$

${\text{[H}}^{+}{\text{] = 10}}^{-6.0}$

${\text{[H}}^{+}\text{] = 1}\times {\text{10}}^{-6}$

Now, let x be the concentration of ${\text{Zn(H}}_{\text{2}}{\text{O)}}_{\text{4}}^{\text{2+}}$.

Equilibrium constant ${\text{(K}}_a)=3.3\times {10}^{-10}$

The formula of equilibrium constant is given by:

${\text{K}}_a=\frac{[{\text{H}}^{\text{+}}][{\text{Zn(H}}_{\text{2}}{\text{O)}}_{\text{3}}{\text{(OH)}}^{\text{+}}]}{{\text{[Zn(H}}_{\text{2}}{\text{O)}}_{\text{4}}^{\text{2+}}]}$

Put the values in above formula.

$3.3\times {10}^{-10}=\frac{(1\times {10}^{-6})(1\times {10}^{-6})}{x-1\times {10}^{-6}}$

Neglect, $\text{1}\times {\text{10}}^{-6}$ in denominator

$3.3\times {10}^{-10}=\frac{(1\times {10}^{-6})(1\times {10}^{-6})}{x}$

$3.3\times {10}^{-10}x=(1\times {10}^{-6})(1\times {10}^{-6})$

$3.3\times {10}^{-10}x=1\times {10}^{-12}$

$x=\frac{1\times {10}^{-12}}{3.3\times {10}^{-10}}=0.003$

Final concentration of U = $1\text{ M}-0.003\text{ M}=0.997\text{ M}$

The relation between half-life and decay constant is given by:

$\lambda \text{=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}$

$\lambda \text{=}\frac{\text{0}\text{.693}}{7\times {10}^8\text{ years}}=9.9\times {10}^{-10}{\text{year}}^{-1}$

The formula for calculating time is given by:

$N_t=N_o\times e^{-\lambda t}$

Where, $N_t$ = number of moles of radioactive element at t time (For uranium = 0.997)

$N_o$ = number of moles of radioactive element at initial time (For uranium = 1.00)

Now, rearrange the formula in terms of time (t)

$t=\frac{1}{\lambda }\ln \left(\frac{N_o}{N_t}\right)$

Put the values.

$t=\frac{1}{9.9\times {10}^{-10}{\text{ year}}^{-1}}\ln \left(\frac{1.00}{0.997}\right)$

$t=\frac{1}{9.9\times {10}^{-10}{\text{ year}}^{-1}}\ln \left(\frac{1.00}{0.997}\right)$

$t=3.03\times {10}^6\text{ year}$

Thus, the 1.00 M solution of U will take $t=3.03\times {10}^6\text{ year}$ to achieve pH = 6.0.

Conclusion

The 1.00 M solution of U will take $t=3.03\times {10}^6\text{ year}$ to achieve pH = 6.0.