How many mL of 0.200 M HCI are needed to neutralize 20.0 mL of 0.150 M $Ba(OH)_2$ ?

Question

24105 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-13T12:30:03

Approx. $-=60*mL$

We use the relationship: $"Moles of stuff"/"Volume of solution"="Molarity"$ .

There are $2xx20xx10^-3*Lxx0.150*mol*L^-1=6.00xx10^-3*mol$ $Ba(OH)_2$ , and this is neutralized according to the following equation:

$Ba(OH)_2(s) + 2HCl(aq)rarrBaCl_2(aq) + 2H_2O(l)$

Given the $"STOICHIOMETRY"$ , we need 2 equiv of hydrochloric acid, and given $0.200*mol*L^-1$ $HCl(aq)$ we need a volume of..........

$(6.00xx10^-3*cancel(mol)xx2)/(0.200*cancel(mol)*L^-1)=0.0600*L$

$-=60*mL$

How many mL of 0.200 M HCI are needed to neutralize 20.0 mL of 0.150 M Ba(OH)_2Ba(OH)_2?