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?
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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