Strong base is dissolved in 625 mL of 0.400 M weak acid (Ka = 3.76 × 10-5) to make a buffer with a pH of 4.09. Assume that the volume remains constant when the base is added.
first calculate pKa from Ka value using
pKa = -logKa = -log[3.76 × 10-5] = 4.4
pH = pka + log (base/acid)
4.09 = 4.4 + log (base/acid)
log (base/acid) = 4.09 - 4.4
log (base/acid) = -0.31
(base/acid) = 10-0.31
(base/acid) = 0.500
moles of the acid = Molarity x volume in liters = 0.4 x 0.625L = 0.25 moles acid
HA + XOH ----> XA + H2O
so here no of moles of salt formed no of moles of strong base
lets base = x
x / 0.25-x = 0.5
x = 0.125 - 0.5x
1.5x = 0.125
x = 0.125 / 1.5
x = 0.083 M
moles = M x volume in liters = 0.083 x 0.625 L = 0.0518 Moles
i calculated both moles and molarity find out whatever you want
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