How many mL of 0.4 M HCl(aq) are needed to neutralize a 187 mL solution of 0.6 M Ba(OH)2?
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first, write a balanced equation for the reaction
Ba(OH)2 + 2HCl --> BaCl2 + 2H2O
we see that 2moles HCl is required to neutralize 1mole of Ba(OH)2 since there are 2 OH- attached.
moles Ba(OH)2 present = 0.187L x 0.6M = 0.112moles
0.112moles Ba(OH)2 requires 0.224moles HCl
0.224moles / 0.4M = volume = 0.561L or 561ml
Ba(OH)2 + 2HCl --> BaCl2 + 2H2O
we see that 2moles HCl is required to neutralize 1mole of Ba(OH)2 since there are 2 OH- attached.
moles Ba(OH)2 present = 0.187L x 0.6M = 0.112moles
0.112moles Ba(OH)2 requires 0.224moles HCl
0.224moles / 0.4M = volume = 0.561L or 561ml