Question

How many milliliters of 0.361 M hydroiodic acid are required to nutralize 13.0 mL of 0.363 M strontium hydroxide according to the UNBALANCED reaction below?

HI (aq) + Sr(OH)2 (aq) → SrI2 (aq) + H2O (l)

Answer 1

The balanced equation is

2 HI (aq) + Sr(OH)2 (aq) → SrI2 (aq) + 2H2O (l)

Number of moles of Sr(OH)2 = molarity * volume of solution in L

Number of moles of Sr(OH)2 = 0.363 * 0.013 = 0.00472 mole

from the balanced equation we can say that

1 mole of sr(OH)2 requires 2 mole of HI so

0.00472 mole of Sr(OH)2 will require

= 0.00472 mole of Sr(OH)2 *(2 mole of HI / 1 mole of Sr(OH)2)

= 0.00944 mole of HI

Molarity of HI = number of moles of Hi / volume of solution in L

0.361 = 0.00944 / volume of solution in L

volume of solution in L = 0.00944 / 0.361 = 0.0261 L

1L = 1000 mL

0.0261 L = 26.1 mL

Therefore, the volume of HI required would be 26.1 mL