Question

A buffer was prepared by dissolving 0.050 mol of the weak acid HA(Ka=1.00*10-5)plus 0.10 mol of its conjugate base K+ A- in 1.0 L calculate the PH?

Answer 1

`pH=5.30`

Explanation:

We use the buffer equation where a weak base, and its conjugate acid are mixed together in appreciable quantities to give a solution that resists gross changes in `pH`...and for such a system...

`pH=pK_a+log_10{[[A^-]]/[[HA]]}`

And so we plug in the noombers...

`pH=+5+underbrace(log_10{[(0.10*mol)/(1.0*L)]/[(0.05*mol)/(1.0*L)]})_(log_10(2)=0.301)`

Because the concentration of the conjugate base is GREATER than that of the parent acid, the `pH` of the buffer tends to towards the basic regime...