How many mL of 2.61 M HBr will be needed to neutralize 183 mL of 1.47 M Mg(OH)2.

Question

How many mL of 2.61 M HBr will be needed to neutralize 183 mL of 1.47 M Mg(OH)2. mL At the equivalence point, moles base (OH-) = moles acid (H+). Each mole of base in this case produces two moles of OH. can someone please explain how to do this one? I tried to do M1V1 = M2V2 and then divide by 2, but Its not working first you have to get the balanced equation of the reaction. then, figure out how many moles of Ba(OH)2 you have (M x L = moles). then figure out how many moles of HBr you need to react completely with the moles of Ba(OH)2. Once you know the moles you can plug moles and molarity into M=mol/L (2.89 = xmol/L) Don?t forget about having to convert tram mL to L a couple 01 times. Let me know if that made any sense.

Explanation / Answer

Hi,

Mg (OH)2 + 2HBr = MgBr2 + 2H2O

at equivalence pont ....
number of equivalents of Mg(OH)2 = number of equivalents of HBr
n1 X M1 X V1 = n2 X M2 X V2

n1 is the n-factor for Mg(OH)2 = 2
M1 is molarity = 1.47 M
V1 is volume of Mg(OH)2 = 183 ml

n2 is n-factor for HBr = 1
M1 is molarity = 2.61 M
V2 is volume of HCl = ?

putting the values ....
2 X 1.47 X 183 = 1 X 2.61 X V2
V2 = 206.13 ml

so 206.13 ml of HBr is needed...

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