How many $sf(OH^-)$ ions are present in 5.0 ml of 0.3M $sf(Ba(OH)_2)$ solution ?

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Answer 1 · 2015-09-26T12:27:18

$1.8xx10^21$

1 mole barium hydroxide produces 2 moles hydroxide ions:

$rarrBa_((aq))^(2+)+2OH_((aq))^-$

$c=n/v$

$n=cxxv$

so no. moles $OH^(-) = 2xx0.3xx0.005=0.003"mol"$

Multiplying by $L$ The Avogadro Constant:

$=6.02xx10^(23)xx0.003=1.8xx10^(21)$ ions.

This ignores the small amount present from the ionisation of water.

How many sf(OH^-)sf(OH^-) ions are present in 5.0 ml of 0.3M sf(Ba(OH)_2)sf(Ba(OH)_2) solution ?