Question

please show work for each part , thank you

part 1

The remaining Points out of 1.00 P Flat question What is the percent ionization of a 0.00473 M solution of HCO H? The K is 1.8x10-4 Multiple tries are permitted; however, 20% (1/5) point will be deducted for each incorrect response

part 2

part 3

part 4

Answer 1

The formula for percent ionization is as follows:-

Percent ionization = (x/C)*100

Where x:- Change in concentration

C:- Original concentration

To find the value of x, we need to draw an ICE table first. Before that we need to write reaction between HCO₂H and water.

HCO₂H (aq) + H₂O (l) -----> HCOO^- (aq) + H₃O⁺ (aq)

	HCO2H	HCOO-	H3O+
I	0.00473	0	0
C	-x	x	x
E	0.00473-x	x	x

The relation for K_a is written as

K_a = [Product]/[Reactant]

1.8 * 10^-4 = x*x/(0.00473-x)

Now apply 5% rule here,

1.8 * 10^-4 =x²/0.00473

Multiply both side by 0.00473

1.8 * 10^-4 * 0.00473 = x²

x² = 8.514 * 10^-7

x² = 85.14 * 10^-1 * 10^-7

x² = 85.14 * 10^-8

Taking square root on both side, we get

x=$\sqrt{85.14*10^{-8}}$

x = 9.2271 * 10^-4

Now put value of x and C in percent ionization,

Percent ionization = (9.2271 * 10^-4 / 0.00473) * 100

= 0.1950 * 100 = 19.50%

Therefore percent ionization of 0.00473 M HCO₂H is 19.50%.

In multiple question we need to solve only one question. If you find any mistake please mention in the comment box.

Thanks.