URGENT!!!!!!! Calculate the concentration of an aqueous solution of Ca(OH)2 that

Question

URGENT!!!!!!!

Calculate the concentration of an aqueous solution of Ca(OH)2 that has a pH of 11.57. **PLEASE SHOW WORK** THANK YOU!!!!!!!!!!

This is my work and the answer I got but it says it is wrong:

pH= 11.57    pOH=14-11.57=2.43

[OH-]= 10^-2.43=3.7*10^-3

I then took 3.7*10^-3 x 2 because there are 2 moles of hydroxide and got 7.4*10^-3(it said this was wrong)

So then I tried putting in 3.7*10^-3 and it still says it is wrong but isn't "deducting points" or taking away attempts (This is on mastering chem)

Explanation / Answer

pH + pOH = 14
pOH = 14 - 11.57 = 2.43

pOH = - log [OH-]
2.43 = - log [OH-]
[OH-] = - antilog (2.43) = 3.7 x 10^-3 M

Ca(OH)2 is a strong base and dissociates completely:
Ca(OH)2(aq) ---------> Ca^2+(aq) + 2OH^--(aq)
This time we will go from right to left:
1.8x10^-3 M ................ 1.8x10^-3 M ...... 3.7x10^-3 M

The concentration of Ca(OH)2 is 1.8x10^-3 M

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