Calculate [OH-] for 1.6×10^−3M of Sr(OH)2?
Calculate [OH-] for 1.6×10^−3M of Sr(OH)2.
I cannot figure out how to do this. My book shows me how to calculate pH for a solution but I am not sure how to find OH. Can someone help me by showing me an equation???
When I attempted I got 7.2*10^-4 and it was wrong.
Also would the equation for the above problem work for these questions:
Calculate [OH-] for 2.210 g of LiOH in 290.0 mL of solution.
Calculate the concentration of an aqueous solution of NaOH that has a pH of 10.78
Any help is appreciated..these are driving me crazy and I cannot make sense out of them- not sure if I am missing or overlooking something.
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Verified answer
Sr(OH)2 --> Sr+2 & 2 (OH-)
so
1.6×10^−3M of Sr(OH)2. releases twice that in (OH-) = 3.2×10^−3M of (OH)-1
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Calculate [OH-] for 2.210 g of LiOH in 290.0 mL of solution
us mlar mass to find moles:
2.210 g of LiOH (1 mole LiOH / 23.95 grams) = 0.09228 moles of LiOH
by the equation:
1 LiOH --> 1 Li & 1 OH-
0.09228 moles of LiOH releases an equal number of moles of (OH)-1 = 0.09228 moles of (OH)-1
find molarity
(0.09228 moles of (OH)-1) / 0.2900 litres = 0.3182 Molar OH-
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since pH & pOH add up to = 14
when the pH = 10.78.
the pOH is 3.22
(OH-) = 10^-3.22
[OH-] = 6.0 X 10^-4 Molar
One mole Sr(OH)2 contains 2 moles of OH Therefore 1.6x10^-3 moles contains 3.2x 10_3moles.
pH +pOH = 1x10^-14 Use this to go back and forth.
Find moles of LiOH first. = 2.210 / 23.9 = 0.0924 moles Molarity = moles/L = 0.0924/0.290 = 0.319
Moles of OH-ions is 0.319 moles /L
NaOH pH = 1078 OH- = 1.65 x10^-11 11.00 - antilog of 22 =10.78. anti log = 1.65
For practise find the pH of 1.65X10^-11.