Highschool Chemistry Question

Question

How many milliliters of 0.108 M KOH must be used neutralize completely 20.0 mL of 0.142 M H3PO4 (aq) solution? 1 H3PO4 (aq) + 3 KOH (aq) → 3 H2O (l) + 1 K3PO4 (aq)

Answer 1

H3PO4 (aq) + 3 KOH (aq) → 3 H2O (l) + 1 K3PO4 (aq)

Convert mL H3PO4 to mol with molarity

20/1000 L (0.142mol)=0.00284 mol H3PO4

convert mol H3PO4 to mol KOH using coefficients

0.00284 mol H3PO4 (3 mol KOH / 1 mol H3PO4)=0.00852 mol KOH

convert mol KOH to L with molarity

0.00852 mol ( 1L / 0.108 mol)=0.0789 L KOH

convert L to mL

0.0789 L (1000mL / 1L)=78.9 mL