Wednesday
July 23, 2014

Homework Help: chem

Posted by Lena on Friday, July 24, 2009 at 1:07am.

Calculate the pH of 0.20 M NaCN solution.

NaCN ---> Na+ + CN-

CN- + H20+ ---> HCN+ + OH-

Initial conc. of CN- = 0.20 mol/L
change = -x
equillibrium = 0.20-x

HCN equill. = +x
OH equill. = 1x10^-7+x

Ka= 6.2 x 10^-10

Kb = KW/Ka = 1x10^-14/6n2 x 10^-10 = 1.6 x 10^-5

1.6 x 10^-5 = [x][1x10^-7+x]/0.2
QUADRATIC FORMULA:
x= 1.8 x 10^-3

pOH = -log(1.8x10^-3)
= 2.73

Ph = 14 - POH
=11.26

DID I DO THIS CORRECTLY?
How would I determine whether or not the CN- was an acid or a base. I was thinking this. The HCN is an acid so when you take the H off of it it becomes the conjugate base. Is this also correct?

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

chem - Which of the following pairs of substances, when mixed in any proportion ...
Chemistry (repost) - Which of the following pairs of substances, when mixed in ...
chem - Which of the following pairs of substances, when mixed in any proportion ...
chemistry - what is the pH of 0.80 M NaCN? what is the pH of 0.80 M NaCN? What ...
chem - Which of the following pairs of substances, when mixed in any proportion ...
Chem - Sodium cyanide NaCN is a solid which melts at 564C to give a liquid that ...
chemistry - A solution of sodium cyanide nacn has a ph of 12.10. How many grams ...
Chemistry - Calculate the pH of the solution that results from mixing 30.0 mL ...
chemistry - A solution of sodium cyanide NACN has a PH of 12.10. How many grams ...
Chemistry - what is the pH of 0.80 M NaCN? What is the concentration of HCN in ...

Search
Members