# chemistry

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Calculate the pH of a 0.10 M NaHX salt solution. Ka (H2X) = 2.1 x 10-2 Ka (HX-) = 7.3 x 10-8

Calculate the pH of a 0.10 M NaHX salt solution. Ka (H2X) = 9 x 10-5 Ka (HX-) = 5.3 x 10-11

• chemistry -

#1 I believe is just the pH of the salt acting as an acid.
..............HX^- ==> H^+ + X^2-
initial.......0.1......0......0
change........-x.......x.......x
equil.......0.1-x......x.......x

Set up k1 and solve for x. You will not be able to solve without using a quadratic equation.

4). This one looks amazingly like NaHCO3 for which (H^+) = sqrt(k1k2)

• chemistry -

DrBob222 don't we have to consider the hydrolysis of HX- here?

H2O(l)+ HX-(aq)<==>H2X(aq)+OH-(aq)-->(1)

equ. (0.1-x) x x
(moldm-3)

Kh= [H2X][OH-]/[HX-]
Kh=kw/ka(HX-)

But we are not given the kW value and we can take it as 14 only if the temperature is 25°C.

Let's assume the temperature is 25°C and kW=14,which gives kh= 10^(-14)/[5.3*(10)^-9]= (1/5.3)*10^(-5) moldm-3=[x^2/(0.1-x)]

if we take (0.1-x) is approximately 0.1,we get,
x^2 = 0.19*(10)^-5==> 2*(10)^-5 M

x==> 4.46*(10)^-2

But we've been given the ka of H2X.

H2X<===> H+(aq) + HX-(aq)->(2)
equ. (x-y) y y
(moldm-3)

which gives,
y^2 =9*(10)^-5 moldm-3 * x
y^2 =9*4.46*(10)^-7 M

Don't we have to consider both (1) and (2) like this to find pH

• chemistry -

x^2==>2*(10)^-6 M
x==>1.41*(10)^-3 M

y^2= 9*1.41*(10)^-8 M
y^2=12.69*(10)^-8 M
y ==>3.56*(10)^-4 M
[H+]=> 3.56*(10)^-4 M

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