How many mL of 0.1566 M HCl solution would be needed to react with 2.88 grams of Ba(OH)2? Then, if a 41.83 mL beaker of 0.2980 M NaOH reacts with 25.00 mL of HNO3 solution, what is the concentration of the HNO3 solution?
no of moles of Ba(OH)2 = W/G.M.Wt
= 2.88/171.34 = 0.0168moles
2HCl(aq) + Ba(OH)2(aq) ----------------> BaCl2(aq) + 2H2O(l)
1 mole of Ba(OH)2 react with 2 moles of HCl
0.0168 moles of Ba(OH)2 react with = 2*0.0168/1 = 0.0336moles of HCl
no of moles of HCl = molarity* volume in L
0.0336 = 0.1566*volume in L
volume in L = 0.0336/0.1566 = 0.215L
the volume of HCl = 215ml >>>>answer
part-B
NaOH(aq) + HNO3(aq) --------------> NaNO3(aq) + H2O(l)
1 mole 1 mole
NaOH HNO3
M1 = 0.2980M M2 =
V1 = 41.83ml V2 = 25ml
n1 = 1 n2 = 1
M1V1/n1 = M2V2/n2
M2 = M1V1n2/V2n1
= 0.2980*41.83*1/(25*1)
= 0.5M
The concentration of the HNO3 solution = 0.5M >>>>answer
How many mL of 0.1566 M HCl solution would be needed to react with 2.88 grams...