Given a diprotic acid, H2A, with two ionization constants of Ka1 = 2.4x10^-4 and Ka2 = 4.9x10^-11, calculate the pH for 0.182 M solution of NaHA.

pH = 6.964 which rounds to 7.0 if 2 sig figs

Explanation:

Ka₁ * Ka₂ = x

2.4x10⁻⁴ * 4.9x10⁻¹¹ = 1.176x10⁻¹⁴

√1.176x10⁻¹⁴ = 1.084435337x10⁻⁷

pH= -log[H⁺]

pH= -log[1.084435337x10⁻⁷]

pH= 6.964796339

h let me think

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer