The acid-dissociation constant at 25.0 c for hypochlorous acid hclo4 is 3.0 10-8 at equilibrium the molarity of h3o in a 0.010 m solution of hclo4 is?

Answer 1

the answer is 2.7x10-2

Start with the reaction (product on top/reactant): Ka=1.0x10-2=[H+][ClO2-] / [HClO2]. We know that we start with 0.100M of HClO2 so we can use that in our equation above. HOWEVER Ka indicates values of a reaction in equilibrium, and the 0.100M is initial value. Construct an ICE table where H+ and ClO2- are both x and the acid value will equal initial minus the change (-x). So what you should have is:

1.0x10-2=x2/0.100-x

now solve using quadratic equation. NOTE: in some cases the x could be negligible however it is not negligible in this problem so you must do the quadratic formula to solve for x. X is your concentration of [H+] = 0.0027=2.7x10-2.

Answer 2

First off, HClO4 is perchloric acid, not hypochlorous. Hypochlorous is HClO. I assume you really mean HClO (hypochlorous) because perchloric is a strong acid and thus virtually completely dissociates in water (has an astronomically high Ka/pKa). You can, in this case, because the Ka is so low (lower than 10^-3 generally), use a simplified equation:

[H+] = sqrt(Acid concentration*Ka)

Therefore:

[H+] = sqrt((0.010 M)(3.0 x 10^-8)) = 0.0000173 M

Which would be a pH of -log(0.0000173) or 4.76. Since it is a weak acid, as you can see, it barely lowers the pH of the solution from ~7.00 to 4.76. If it were a strong acid (say HCl or HI), it's pH would be -log(.010) or 2.00. Because pH is on a logarithmic scale, this is over 100 times more acidic than the weak acid is at the same concentration.

Answer 3

Number of moles of HOCl = 0.1 M * 0.04 L

= 0.004 mol

After addition of 10.0 mL NaOH

Number of moles of NaOH = 0.1 M * 0.01 L

= 0.001 mol

HOCl + NaOH → NaOCl + H2O

0.001 mol 0.001 mol 0.001 mol

The remaining HOCl and NaOCl form the buffer system

The number of moles of remaining HOCl = 0.004 -0.001 = 0.003 mol

pH = pKa + log [NaOCl] /[HOCl]

= -log 3.0 x 10-8 + log 0.001 mol /0.003 mol

= 7.522 -0.477

= 7.044

2. After the addition of 20.0 mL of NaOH

Number of moles of NaOH = 0.1 M * 0.02 L = 0.002 mol

HOCl + NaOH → NaOCl + H2O

0.002 mol 0.002 mol 0.002 mol

The remaining HOCl and NaOCl form the buffer system

The number of moles of remaining HOCl = 0.004 -0.002 = 0.002 mol

pH = pKa + log [NaOCl] /[HOCl]

= -log 3.0 x 10-8 + log 0.022 mol /0.022 mol

= 7.522

3 . After the addition of 30 mL of NaOH

Number of moles of NaOH = 0.1 M * 0.03 L = 0.003 mol

HOCl + NaOH → NaOCl + H2O

0.003 mol 0.003 mol 0.003 mol

The remaining HOCl and NaOCl form the buffer system

The number of moles of remaining HOCl = 0.004 -0.003 = 0.001 mol

pH = pKa + log [NaOCl] /[HOCl]

= -log 3.0 x 10-8 + log 0.003 mol /0.001 mol

= 7.522 +0.477

= 7.99

2. After the addition of 20.0 mL of NaOH

Number of moles of NaOH = 0.1 M * 0.02 L = 0.002 mol

HOCl + NaOH → NaOCl + H2O

0.002 mol 0.002 mol 0.002 mol

The remaining HOCl and NaOCl form the buffer system

The number of moles of remaining HOCl = 0.004 -0.002 = 0.002 mol

pH = pKa + log [NaOCl] /[HOCl]

= -log 3.0 x 10-8 + log 0.002 mol /0.002 mol

= 7.522

Answer 4

A constant is not needed because HClO4 is a strong acid and will fully dissociate in solution.