What volume in milliliters, of 0.35 m ca(oh)2 is needed to completely neutralize 278.3 ml of a 0.32 m hi solution?

Volume Ca(OH)2 =  mL ?

Molarity = 0.35 M

Volume HI =  278.3 mL / 1000 => 0.2783 L 

Molarity HI = 0.32 M

number of moles HI :

n = M x V

n = 0.32 x 0.2783

n = 0.089056 moles of HI

mole ratio :

Ca(OH)2 + 2 HI = CaI2 + 2 H2O

1 mole Ca(OH)2 2 moles HI
moles Ca(OH)2  0.089056 moles of HI

moles Ca(OH)2 = 0.089056 x 1 / 2

moles Ca(OH)2 = 0.089056 / 2

= 0.044528 moles of Ca(OH)2

Volume of Ca(OH)2 :

M = n / V

0.35 = 0.044528 / V

V = 0.044528 / 0.35

V = 0.127 L in mL :

0.127 x 1000 = 127 mL 

hope this helps!

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