4.48
pH=pKa+log([A-/HA])
25% deprotonated tells us that A- is .25 and that the rest (75% is protonated) thats .75.
4 = pKa + log \frac{.25}{.75}
4 - log \frac{.25}{.75} = pKa
4.48=pKa
The ion is Al³⁺
mass number - number of neutrons= atomic number
27 - 14 = 13
Aluminum has an atomic number of 13, thus we know this is the metal in question. Also, because the aluminum has only 10 electrons, (3 less than a neutral atom of aluminum would have), its charge must be 3+
Answer:
NH3(aq) + HNO3(aq) → NH4NO3(aq) Calculate the volume of an acid (1.5 M HNO3) needed to neutralize the 1.5 M HNO3.
Explanation:
