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Answer 1 · 2015-10-28T17:42:28

$K_b=1.8xx10^(-10)color(white)(x)"mol/l"$

We can firstly find the $pK_a$ value using this expression for a weak acid:

$pH=1/2(pK_a-logc)$

Where $c$ is the concentration of the acid.

$2.5=1/2pK_a-1/2log(0.18)$

$2.5=1/2pK_a+0.372$

$1/2pK_a=2.128$

$pK_a=4.256$

We now use the relationship:

$pK_a+pK_b=14$

$pK_b=14-4.256=9.744$

This means that $-logK_b=9.744$

From which $K_b=1.8xx10^(-10)color(white)(x)"mol/l"$