fin the ph for buffer 0.12m,0.16m NaN02 with the given ka(HNO2)= .5x10^-
5
You need to learn, and use, the correct notation for concentration. When you write 0.12m you say 0.12 molal where molal is mols/kg solvent. If you meant to say 0.12 molar where molar is mols/L of solution you should write 0.12 M. There is a difference between 0.12 m and 0.12 M.
I am assuming you meant 0.12 M HNO2 (you didn't type the HNO2) and 0.16 M NaNO2. HNO2 is the acid; NaNO2 is the base. Use the Henderson-Hasselalch equation of
pH = pKa + log [(base)/(acid)]
Substitute and solve. Post your work if you get stuck.
To find the pH of a buffer solution, we can use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log([A-]/[HA])
where pH is the measure of acidity or basicity, pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, the acid is HNO2, and the conjugate base is NO2-. The acid dissociation constant (Ka) for HNO2 is given as 5x10^-5.
We are given the concentrations of the acid (HNO2) and the conjugate base (NO2-), which are 0.12 M and 0.16 M, respectively.
First, we need to find pKa from Ka. Since pKa is the negative logarithm of Ka, we have:
pKa = -log(Ka)
pKa = -log(5x10^-5)
pKa = 5
Now, we can substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 5 + log(0.16/0.12)
pH = 5 + log(1.333)
pH ≈ 5 + 0.125
pH ≈ 5.125
Therefore, the pH of the buffer solution containing 0.12 M HNO2 and 0.16 M NaN02, with a pKa of 5x10^-5, is approximately 5.125.
To find the pH of a buffer solution, you need to consider the dissociation of the weak acid and its conjugate base.
In this case, the weak acid is HNO2 (nitrous acid) and its conjugate base is NO2- (nitrite ion). The given Ka value for HNO2 is 5x10^(-5).
To solve this problem, you can use the Henderson-Hasselbalch equation, which is given as:
pH = pKa + log ([base] / [acid])
where pH is the negative logarithm of the hydrogen ion concentration, pKa is the negative logarithm of the acid dissociation constant, [base] is the concentration of the conjugate base, and [acid] is the concentration of the weak acid.
In this case, the concentration of the weak acid (HNO2) is not given, but we know that the total concentration of the buffer solution is 0.12 M + 0.16 M = 0.28 M. Since HNO2 and NaN02 are in a 1:1 ratio, it means the concentration of HNO2 is also 0.12 M.
Now, let's plug the values into the Henderson-Hasselbalch equation:
pH = -log(5x10^(-5)) + log(0.16/0.12)
Calculating the logarithms:
pH = -(-4.3) + log(1.33)
pH = 4.3 + 0.12
pH ≈ 4.42
Therefore, the pH of the buffer solution is approximately 4.42.