(Unionized, weak acid and common ion effect)

$HA + NaOH \to NaA + {H_2}O$

$NaA \to N{a^ + } + {A^ - }$      (ionized)

${K_a} = \frac{{[{H^ + }][{A^ - }]}}{{[HA]}}$

Given,  $pH = 6,\,[{H^ + }] = 1 \times {10^{ - 6}}$

$[{H^ + }] = \frac{{{K_a}[Acid]}}{{[Salt]}}$

$\frac{{[Salt]}}{{[Acid]}} = \frac{{{K_a}}}{{[{H^ + }]}} = \frac{{{{10}^{ - 5}}}}{{{{10}^{ - 6}}}} = 10:1$