# Question 37f75

Feb 23, 2017

$2 \left(1082403910671456 \cdot {10}^{15}\right) + 2 \left(300790362857502000000000\right)$= calculate yourself the number of ions

#### Explanation:

Solubility of $C a S {O}_{4} = \text{ 0.21g"/"100ml}$

Amount of water required to dissolve 68g of $C a S {O}_{4}$

=$\text{100ml"/"0.21g} \cdot 68 g = 32380.952380952380952380952380952 L$

This means there is 32380.9524L of water and 68g of $C a S {O}_{4}$

No. of ions in 68g of calcium sulphate in water

No. of moles in 68g of $C a S {O}_{4}$

$\text{68g"/"136.14g/mol" = 0.49948582341"moles}$

$C a S {O}_{4} r i g h t \le f t h a r p \infty n s C {a}^{+ 2} + S O {4}^{2} -$

No. of $C {a}^{+ 2}$ ions

$0.49948582341 \text{moles" * 1/1 * "6.022 * 10^23Ca(+2)ions"/"1mol Ca}$

$= 300790362857502000000000 \text{ions of Ca}$

This there are also 300790362857502000000000 ions of $S O {4}^{-} 2$

No. of ions in 32380.952380952380952380952380952L

"32380.952380952380952380952380952L * 1000g = "32380952.380952380952380952380952g = 32380952.4g("rounded off)#

Moles = 32380952.4g/18.01528g

= 1 797415.99 moles

${H}_{2} O r i g h t \le f t h a r p \infty n s {H}^{+} + O {H}^{-}$

Number of H^+ ions = OH^- ions

$\text{1797415.99248mol" * "1mol H+"/"1molH2O" * (6.023 * 10^23"mol" H^+ "mol")/(1 "mol"H_2) = 1082403910671456000000000000000"ions}$

total ions

$2 \left(1082403910671456 \cdot {10}^{15}\right) + 2 \left(300790362857502000000000\right)$= calculate yourself