chemistry
posted by Anonymous on .
Calculate the pH of a 0.10 M NaHX salt solution. Ka (H2X) = 2.1 x 102 Ka (HX) = 7.3 x 108
Calculate the pH of a 0.10 M NaHX salt solution. Ka (H2X) = 9 x 105 Ka (HX) = 5.3 x 1011

#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.......0.1x......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) 
DrBob222 don't we have to consider the hydrolysis of HX here?
H2O(l)+ HX(aq)<==>H2X(aq)+OH(aq)>(1)
equ. (0.1x) x x
(moldm3)
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) moldm3=[x^2/(0.1x)]
if we take (0.1x) 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. (xy) y y
(moldm3)
which gives,
y^2 =9*(10)^5 moldm3 * x
y^2 =9*4.46*(10)^7 M
Don't we have to consider both (1) and (2) like this to find pH 
Some corrections should be added!
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