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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: nooshnoosh95 on May 14, 2012, 10:19:09 pm

Title: chem questions
Post by: nooshnoosh95 on May 14, 2012, 10:19:09 pm

help please
300 mL of a 0.35M Ca(OH)2 solution was reacted with 300 mL of 0.4M HCl. the pH of the resultant solution is closest to:

A. 0.82
B. 1.12
C. 12.88
D. 13.18

i thought it was C but the answer is D...??

Title: Re: chem questions
Post by: charmanderp on May 14, 2012, 10:41:55 pm

n(Ca(OH)2)=0.3*0.35=0.105mol
n(OH-)=0.105*2=0.210mol
n(HCL)=0.3*0.4=1.2mol=n(H+)

n(OH-)-n(H+)=0.09mol

C=0.09/0.6=0.15M
pOH=-log(0.125)=0.82
pH=14-pOH=13.18

Which is slightly different to your options due to rounding of numbers. Calculating it by drawing up an equation and finding the limiting reagent, how much base in excess, etc, gave me the same result.

EDIT: Nope, not because of rounding, rather because I am blind and read 0.4 as 0.45 :/

Title: Re: chem questions
Post by: DisaFear on May 14, 2012, 10:42:25 pm

n((OH^-)₂) = 2x0.35x0.3 = 0.21 mol (Due to the (OH)₂, 2 of them dudes)
n(H⁺) = 0.4x0.3 = 0.12 mol

What is left in excess after they react? 0.09 mol of the base





[OH^-] = 0.15g/mol

[OH^-]x[H₃O⁺] = 10^-14

[H₃O⁺] = 10^-14/0.15 = 6.667x10^-14

-log(6.667x10^-14) = 13.176

Title: Re: chem questions
Post by: doh07 on May 14, 2012, 10:42:45 pm

got D. mol (OH) = 0.21
mol (H+) = 0.12
subtract and u get 0.09 mol OH- left over.
C(OH-) = n/v = .09/.6 = .15 M
plug it into 10^-14/.15 and u get some number which u sub into -log(ans).
and u get D!

Title: Re: chem questions
Post by: nooshnoosh95 on May 14, 2012, 10:48:04 pm

ohh thanks babes forgot to multiply by 2 for the mols of Ca(OH)2