N = PV = (190 atm)(35 L) = 260 moles of gas RT (0.0821 L.atm/mol.K)(315 K)
Answer:
0.001 M OH-
Explanation:
[OH-] = 10^-pOH, so
pOH + pH = 14 and 14 - pH = pOH
14 - 11 = 3
[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-
Hey there!
Values Ka1 and Ka2 :
Ka1 => 8.0*10⁻⁵
Ka2 => 1.6*10⁻¹²
H2A + H2O -------> H3O⁺ + HA⁻
Ka2 is very less so I am not considering that dissociation.
Now Ka = 8.0*10⁻⁵ = [H3O⁺] [HA⁻] / [H2A]
lets concentration of H3O⁺ = X then above equation will be
8.0*10−5 = [x] [x] / [0.28 -x
8.0*10−5 = x² / [0.28 -x ]
x² + 8.0*10⁻⁵x - 2.24 * 10⁻⁵
solve the quardratic equation
X =0.004693 M
pH = -log[H⁺]
pH = - log [ 0.004693 ]
pH = 2.3285
Hope that helps!
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