How many grams of sulphuric acid is to be dissolved to prepare 200 mL aqueous solution having concentration of $\left\lbrack H_{3}O^{+} \right\rbrack$ ions 1 M at 25$℃$ temperature. $\left\lbrack H = 1,\ O = 16,\ S = 32\ g.{mol}^{- 1} \right\rbrack$

Question

Dr. Abubacker · Accepted Answer

(c)
Molarity, M = $\frac{W_{2} 	imes 1000}{M_{2}\  	imes Vol.(mL)};$ where $w_{2}$ mass of $H_{2}{SO}_{4}$ in g, $M_{2}$ is the molar mass of $H_{2}{SO}_{4}$
$w_{2} = \frac{1\  	imes 98\  	imes 200\ }{1000} = 19.6\ g$
$H_{2}{SO}_{4} + \ 2H_{2}O\  ightleftharpoons {2H}_{3}O^{+ \ } + {SO}_{4}^{2 -}$
But according to equation 1 mole of $H_{2}{SO}_{4}$ gives 2 mole of [$H_{2}O^{+ \ }$] ions. So, the amount of $H_{2}{SO}_{4}$ to prepare 200 mL solution having the 1 M concentration o$H_{3}O^{+ \ }$ ions is 19.6/2 = 9.8 g.

Q. How many grams of sulphuric acid is to be dissolved to prepare 200 mL aqueous solution having concentration of $\left\lbrack H_{3}O^{+} \right\rbrack$ ions 1 M at 25$℃$ temperature. $\left\lbrack H = 1,\ O = 16,\ S = 32\ g.{mol}^{- 1} \right\rbrack$

Explanation

Reviewed by Dr. Abubacker

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