We are given with the equilibrium constant of acid, HF and is asked to calculate the pH of 0.30 M NaF solution. The formula to be followed is
Ka = [H+][F-]/[HF]Ka = 7.2 x 10 -4 = x^2/[0.3-x]x = [H+]= pH = -log (H+) = 1.84
Answer:
Learn from their experiment and know the possible results from trial and error
Answer:
8o/58*6*10power23 gives u the answer. molecules=moles*avagadrono
moles=weight/molecular wt
hope u understandmark me as branliest
A neutral atom of potassium has 19 electrons.
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:
PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32
