Calculate the volume of 0.5 molar HCl required to titrate 22ml of 0.5 molar Ba(OH)2
To calculate the volume of 0.5 Molar HCl required to titrate 22 ml of 0.5 Molar Ba(OH)2, we can use the concept of stoichiometry and the balanced chemical equation of the reaction between HCl and Ba(OH)2.
The balanced chemical equation for the reaction between HCl and Ba(OH)2 is:
2 HCl + Ba(OH)2 -> 2 H2O + BaCl2
From the equation, we can see that 2 moles of HCl react with 1 mole of Ba(OH)2 to produce 2 moles of water and 1 mole of BaCl2.
To find the volume of HCl required, we need to determine the number of moles of Ba(OH)2 present in 22 ml of 0.5 Molar Ba(OH)2 solution.
The number of moles can be calculated using the formula:
moles = concentration (Molarity) x volume (in liters)
Converting 22 ml to liters, we divide by 1000:
volume (in liters) = 22 ml / 1000 = 0.022 L
Now, we can calculate the number of moles of Ba(OH)2:
moles of Ba(OH)2 = concentration x volume = 0.5 M x 0.022 L = 0.011 moles
Since the stoichiometry of the reaction is 2:1 for HCl and Ba(OH)2, this means that we need twice as many moles of HCl as moles of Ba(OH)2:
moles of HCl required = 2 x moles of Ba(OH)2 = 2 x 0.011 moles = 0.022 moles
To find the volume of HCl required, we can use the formula:
volume (in liters) = moles / concentration (Molarity)
volume of HCl required = 0.022 moles / 0.5 M = 0.044 L
Converting this volume to milliliters, we multiply by 1000:
volume of HCl required = 0.044 L x 1000 = 44 ml
Therefore, the volume of 0.5 M HCl required to titrate 22 ml of 0.5 M Ba(OH)2 is 44 ml.