# Chemistry

posted by .

In a 1.760 M aqueous solution of a monoprotic acid, 3.21% of the acid is ionized. What is the value of it's Ka?

x=1.760M * 3.21/100 - .0565M

Ka expression is Ka= {[H+][A-]} / [HA]

• Chemistry -

Did you make a typo; that = 0.0565M and not -.
All of that looks good to me.

## Similar Questions

1. ### Chemistry

A 0.040M solution of a monoprotic acid is 14% ionized. Calculate the Ka for the weak acid. ---------------- so: HX <---> H+ + X- Ka= [H+][X-]/ [HX] Since its a monoprotic acis i know the concentration of H and X will be equal …
2. ### Chemistry

A 0.100 mol quantity of a monoprotic acid HA is added to 1.00 L of pure water. When equilibrium is reached, the pH of the solution is 3.75. What is the value of Ka for the acid HA?
3. ### Chemistry

A 0.490 M solution of an unknown monoprotic weak acid, HA, is 2.80% ionized. What is the value of Ka for HA?
4. ### Chemistry

The pH of a 0.010 M aqueous solution of a weak monoprotic acid, HX, is 4.5. What is the value of the acid ionization constant?
5. ### Chemistry

What volume of 0.200M of aqueous solution of formic acid, a weak monoprotic acid (KA = 1.78x10-4) and 0.200M aqueous solution of NaOH would you mix to prepare a 500mL of a buffer solution of pH = 4.0.
6. ### chemistry

In a 1.070 M aqueous solution of a monoprotic acid, 4.81% of the acid is ionized. What is the value of its Ka?
7. ### Chemistry 104

A 0.30M solution of a weak monoprotic acid is 0.41% ionized. what is the acid-ionization constant, Ka for this acid?