0.1 mole of ch3nh2(kb=5*10^-4) is mixed with 0.08 mole of hcl and diluted to one litre the [h+] in solution is
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By Henderson-Hasselbalch Equation:
=> pH = pKa + log (base/acid)
pKb = - log 5 x 10⁻⁴
= 3.30103
pKa = 14 - 3.30103
= 10.69897
pH = 10.69897 + log (0.02 / 0.08)
pH = 10.69897 + (-0.60206)
pH = 10.09691
Antilog the pH to get the H+ conc. which will be = 8 x 10⁻¹¹ M
_____________________________________________
*******
By Henderson-Hasselbalch Equation:
=> pH = pKa + log (base/acid)
pKb = - log 5 x 10⁻⁴
= 3.30103
pKa = 14 - 3.30103
= 10.69897
pH = 10.69897 + log (0.02 / 0.08)
pH = 10.69897 + (-0.60206)
pH = 10.09691
Antilog the pH to get the H+ conc. which will be = 8 x 10⁻¹¹ M
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