Stoichiometry to find concentration and pH

[pleace]

Homework Statement

15.1g of calcium hydroxide completely dissociates in distilled water forming 855ml of basic
solution.
a) Dissociation Equation
b) Calculate the concentration of OH in the solution
c)Calculate the pH of the above solution

Homework Equations

c=n/v
pH=-log[H]

The Attempt at a Solution

a) Ca(OH)2 (s) = Ca2+ + 2OH -
i think that equation is right?
b) using molar mass ofCa(OH)2 and 15.1g, the moles of Ca(OH)2 is around 0.2moles.
since Ca(OH)2 and OH are in 1:2 ratio, then there is 0.4 moles Oh.
c=n/v
c=0.4/0.855 =0.452 mol/L(DO i use 0.855L for the volume even tho that includes the entire basic solution which is Ca2+ + 2OH - ?)
c) Now this is the part where i am really unsure about.
pH = -log[H], but since i want to find the entire solution's pH, what concentration do u use? do i divide 0.452/2=0.226
and then go -log0.226 = 0.64.

Thank you so much if you try to help me :)

 

[Saitama]

\
c=0.4/0.855 =0.452 mol/L(DO i use 0.855L for the volume even tho that includes the entire basic solution which is Ca2+ + 2OH - ?)

The volume is 0.855 mL as per the question. Are you sure it's 0.855 mL in the question? I seem to get some weird answers if i use 0.855 mL. But, i will check out again.

Now this is the part where i am really unsure about.
pH = -log[H], but since i want to find the entire solution's pH, what concentration do u use? do i divide 0.452/2=0.226

We do not have the concentration of H+. Do you know any relation between p[H] and p[OH]?

 

[pleace]

i meant to put [OH] i forgot to type the O. and yes i am 100% sure that it says 855ml

 

[Infinitum]

Hi pleace!

pH = -log[H], but since i want to find the entire solution's pH, what concentration do u use? do i divide 0.452/2=0.226

No, you are halving the actual concentration that way...you need to pH of the solution which is defined to be the concentration of H+ in the solution. So use it to find pOH and then convert it to pH.

 

[Saitama]

I misread the question, its 855 mL and i am seeing it as 0.855 mL, i am such a dumb. :tongue:

Anyways, back to question, do you know about the relation: p(OH)+p(H)=14?
Find pOH and use the relation to find pH.

 

[HeavyMetal]

Homework Statement

15.1g of calcium hydroxide completely dissociates in distilled water forming 855ml of basic
solution.
a) Dissociation Equation
b) Calculate the concentration of OH in the solution
c)Calculate the pH of the above solution

Homework Equations

c=n/v
pH=-log[H]

The Attempt at a Solution

a) Ca(OH)2 (s) = Ca2+ + 2OH -
i think that equation is right?
b) using molar mass ofCa(OH)2 and 15.1g, the moles of Ca(OH)2 is around 0.2moles.
since Ca(OH)2 and OH are in 1:2 ratio, then there is 0.4 moles Oh.
c=n/v
c=0.4/0.855 =0.452 mol/L(DO i use 0.855L for the volume even tho that includes the entire basic solution which is Ca2+ + 2OH - ?)
c) Now this is the part where i am really unsure about.
pH = -log[H], but since i want to find the entire solution's pH, what concentration do u use? do i divide 0.452/2=0.226
and then go -log0.226 = 0.64.

Thank you so much if you try to help me :)

A is right, and B is right too. Concentration should be measured in molarity which you correctly represented as moles/liters. But don't round your sig figs so early, this could be bad.

1. Calculate the molar mass: this is about 74.0932 g/mol.
2. Multiply that by your 15.1 grams to get 0.2037973 mol.
3. Multiply that by two to get 0.4075946 mol.
4. Divide by .855 L to get a 0.476719 M solution of hydroxide OH^- ions. But to get the right answer for C, you must take into account that we only have 3 sig figs, so the concentration would be 0.477 M.
5. Employ the rule pH + pOH = 14. In other words, pH = 14 - pOH. pOH = -log[OH^-]. Our concentration is 0.476719 M, so we just take the negative log of that. -log[0.476719] = 0.740828. Therefore, pH = 14 - 0.740828 = 13.259172.
6. LAST STEP: Sig figs. The equation you gave us has two numbers: 855mL, and 15.1g. They both contain three sig figs. But when we use logs, the only sig figs are the ones to the right of the decimal. So 13.259172 pH becomes 13.259 pH.

EDIT:
This is why sig figs are so important. If we used your numbers, we would get a concentration of about 0.452 M, the total pH would be 13.655 instead of 13.259. That's about 0.396 off.