Calculate the ph of 0.25mol h2so4
ka= 1.9 x 10-5
only have experience with simpler ones like koh, the h2so4 is whats confusing me

This problem uses the relationship between Ka and the concentrations of the ions. Calculations are as follows:

1.9 x 10-5= x^2 / (0.25 - x)

x is very small and the denominator is approximately equal to 0.25. Thus, x is 2.2 x 10^-3

pH = -log (2.2 x 10^-3) = 2.66

these nuts

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.