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What is the value of [OH-] in a 0.0800 M solution of Sr(OH)2? What is the...

What is the value of [OH-] in a 0.0800 M solution of Sr(OH)2? What is the concentration of OH- ions after adding 12 mL of 0.465 M HCl to 100 mL of 0.0800 M Sr(OH)2?

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Homework Answers

Answer #1

1 mole of Sr(OH)2 will give two mole of OH-

Therefore

0.0800 M Sr(OH)2 will give 0.1600    M OH-

The balance recation between HCl and Sr(OH)2 is as follows:

Sr(OH)2 +2HCl = SrCl2 +2 H2O

Mole of HCl = molarity * volume in L

= 0.465*12/1000

= 0.00558 Mole HCl

Mole of Sr(OH)2 = 0.0800*100/1000

= 0.008 mole Sr(OH)2

Now calculate the mole Sr(OH)2 as follows:

0.00558 Mole HCl *1 mole Sr(OH)2 /2 mole HCl

= 0.00279 mole Sr(OH)2

Reaming mole Sr(OH)2 =total moles of Sr(OH)2- used

= 0.008 -0.00279 mole Sr(OH)2

= 0.00521 mole Sr(OH)2

Total volume =100 ml+12 ml

= 112 ml

=0.112 ml

Molarity Sr(OH)2 after reaction = number of mole s/ volume in l

= 0.00521 mole Sr(OH)2 0/0.112

= 0.0465 M Sr(OH)2

1 mole of Sr(OH)2 will give two mole of OH-

Therefore

0.0465 M Sr(OH)2 will give 0.0930            M OH-

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