chem questions

[nooshnoosh95]

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...??

[charmanderp]

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 :/

[DisaFear]

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

[doh07]

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!

[nooshnoosh95]

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