4.18

vetykloridin konsentraatio

$\frac{1{,}65\ g}{36{,}46\ \frac{g}{mol}}=0{,}045255...\ \frac{mol}{l}$

$c\left(NaOH\right)=0{,}350\ \frac{mol}{l}$

reaktioyhtälö

$NaOH\left(aq\right)+HCl\left(aq\right)\rightarrow NaCl\left(aq\right)+H_2O\left(l\right)$

vetykloridin määrä

$0{,}045255...\ \frac{mol}{l}\cdot0{,}004l=0{,}0001810...mol$

natriumhydroksidia tarvitaan saman verran

$V=\frac{n}{c}=\frac{0{,}0001810...mol}{0{,}350\ \frac{mol}{l}}=0{,}00051720...l\approx5{,}17ml$
b)

$c\left(H_2SO_4\right)=\frac{10ml\cdot1{,}25\ \frac{mol}{l}}{100ml}=0{,}125\ \frac{mol}{l}$
$n\left(H_2SO_4\right)=0{,}025l\cdot0{,}125\ \frac{mol}{l}=0{,}003125\ mol$

$H_2SO_4\left(aq\right)+2\ NaOH\left(aq\right)\rightarrow Na_2SO_4\left(aq\right)+2\ H_2O\left(l\right)$

NaOH tarvitaan tuplasti

$2\cdot n\left(H_2SO_4\right)=0{,}00625mol$

$V=\frac{n}{c}=\frac{0{,}00625\ mol}{0{,}350\ \frac{mol}{l}}=0{,}017857...l\approx17{,}86ml$