What is the (H3O+) and

the ph of a buffer that consists of 0.25M HNO2 and 0.69M KNO2?

(Ka of HNO2 = 7.1 x 10^-4)

is this right?
pH of acidic buffer

pH = pka + log(salt/acid)

pka = -log ka = -log(7.1*10^(-4)) = 3.15

= 3.15+log(0.25/0.69)

pH = 2.71

[H3O+] = 10^-pH = 10^(-2.71) = 0.001945 M
Thank you.

In the HH equation didn't you say it was salt/acid. KNO2 is the salt; HNO2 is the acid.

[base] > [acid] so your pH should be larger than pka. check DrBob's answer :)

Your equation for calculating the pH of an acidic buffer is correct. However, there is a mistake in your calculation.

pKa = -log(Ka) = -log(7.1 x 10^-4) = 3.15

pH = pKa + log(salt/acid) = 3.15 + log(0.69/0.25)
= 3.15 + log(2.76)
= 3.15 + 0.44
= 3.59

So, the pH of the buffer is 3.59, not 2.71.

To calculate [H3O+], you are correct in using the formula [H3O+] = 10^-pH. Thus,

[H3O+] = 10^(-3.59) = 0.000381 M

So, the [H3O+] concentration in the buffer is 0.000381 M, not 0.001945 M.

Therefore, the correct pH of the buffer is 3.59, and the [H3O+] concentration is 0.000381 M.

To find the pH of the buffer, you can use the Henderson-Hasselbalch equation:

pH = pKa + log(salt/acid)

First, calculate the pKa by taking the negative logarithm of the Ka value:

pKa = -log(7.1 x 10^-4) = 3.15

Next, substitute the concentrations of the salt (KNO2) and acid (HNO2) into the equation:

pH = 3.15 + log(0.69/0.25)

Calculate this to get the pH value:

pH = 3.15 + log(2.76)

Finally, evaluate the logarithm of the ratio to find the pH:

pH ≈ 3.15 + 0.44 ≈ 3.59

So, the pH of the buffer is approximately 3.59, not 2.71 as you mentioned.

To find the concentration of H3O+, you can use the inverse of the pH value:

[H3O+] = 10^(-pH)

Substitute the pH value into the equation:

[H3O+] = 10^(-3.59)

Calculate this to find the concentration:

[H3O+] ≈ 0.0021 M (rounded to 4 significant figures)

Therefore, the concentration of H3O+ in the buffer is approximately 0.0021 M, not 0.001945 M as you mentioned.

Please note that the values mentioned here have been approximated for simplicity and may have more accurate values depending on the significant figures provided in the original question or any specific rounding requirements.