Question

Close Problens Calculate the pH of a weak acid solution (IHAb 100.Ka). Calculate the pH of a 0.406 M aqueous solution of nitrous acid (HNO,, K-4.5x10) and the equilibrium concentrations of the weak acid and its conjugate base. pHH [HNO2 lequilibrium NO2 lequilbrium Show Approach

Help plz

Answer 1

Given

[HNO₂] = 0.406 M

To calculate pH of a weak acid we have to use ICE table

	HNO2	H2O <===>	H3O+	NO2-
I	0.406 M	-	0	0
C	-x	-	+x	+x
E	0.406-x	-	x	x

Ka of HNO2 = 4.5 x 10^-4

but

$K_a =\frac{[H_3O^+][NO_2^-]}{[HNO_2]}$

$4.5\times 10^{-4} =\frac{[x][x]}{[0.406-x]}$

$4.5\times 10^{-4} =\frac{[x]^2}{[0.406-x]}$

Neglecting denominator x (0.406-x = 0.406)

$4.5\times 10^{-4} =\frac{[x]^2}{[0.406]}$

$4.5\times 10^{-4}\times 0.406 =[x]^2$

$1.827\times 10^{-4} =[x]^2$

$x=\sqrt{1.827\times 10^{-4}}$

x = 0.0135 M

[H₃O⁺] = x = 0.0135 M

[NO₂^-] = x= 0.0135 M

[HNO₂] = 0406 - x = 0.406- 0.0135 = 0.392 M

Formula

pH = -log[H₃O⁺]

pH = -log (0.0135)

pH = 1.87

------------------------------------------

Given,

[HC₉H₇O₄] = 0.0170 M

Using ICE table to solve for pH

	HC9H7O4	H2O <==>	C9H7O4-	H3O+
I	0.0170 M	-	0	0
C	-x	-	+x	+x
E	0.0170 -x	-	x	x

Given

$K_a = \frac{[C_9H_7O_4^-][H^+]}{[HC_9H_7O_4]}$

$3.4\times 10^{-4}= \frac{[x][x]}{[0.0170-x]}$

$3.4\times 10^{-4}\times [0.0170-x]= [x]^2$

here we are asked to solve by quadratic equation so we cannot neglect x

$x^2=5.78\times 10^{-6}-3.4\times 10^{-4}(x)$

$x^2+3.4\times 10^{-4}(x)+(-5.78\times 10^{-6})=0$

it is in the form of ax² +bx +c =0

formula to solve for x

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x = \frac{(-3.4\times 10^{-4})\pm \sqrt{(3.4\times 10^{-4})^2-4\times (-5.78\times 10^{-6})}}{2}$ $x = \frac{(-3.4\times 10^{-4})\pm \sqrt{2.32\times 10^{-5}}}{2}$

$x = \frac{(-3.4\times 10^{-4})\pm 0.00482}{2}$

$x = \frac{(-3.4\times 10^{-4})+ 0.00482}{2}, x = \frac{(-3.4\times 10^{-4})- 0.00482}{2}$

Concentration wont be in negative

so X= 0.00224 M

[H₃O⁺] = 0.00224 M

[C₉H₇O₄^-] = x =  0.00224 M

[HC₉H₇O₄] = 0.0170- x = 0.0170- 0.00224 = 0.0148 M

pH = -log[H₃O⁺]

pH = -log(0.00224)

pH = 2.65