# chemistry

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calculate the Ph of 0.3 M sodium propanoate, Ka 2.0 x 10 base -3. then calculate the pH after 0.03 mol NCL is added to 2L of the first solution.

• chemistry -

Sodium propanoate we will call NaP. It's the propanoate ion that is hydrolyzed in solution.
.......P^- + HOH ==> HP + OH^-
I.....0.3............0....0
C......-x............x....x
E.....0.3-x..........x....x

Kb for P^- = (Kw/Ka for HP) = (x)(x)/(0.3-x)
Solve for x = (OH^-) and convert to pH.

For the second part, is NCl a typo for NaCl? Whether yes or no, write the reaction between HP and NCL (or whatever).

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