What volume of 5.00 × 10-3 m hno3 is needed to titrate 80.00 ml of 5.00 × 10-3 m ca(oh)2 to the equivalence point? what volume of 5.00 × 10-3 m hno3 is needed to titrate 80.00 ml of 5.00 × 10-3 m ca(oh)2 to the equivalence point? 80.0 ml 160. ml 10.0 ml 40.0 ml?

= 160 mL

Explanation:

The equation for the reaction is;

Ca(OH)2 + 2HNO3 > Ca(NO3)2 + 2H2O

Moles of Ca(OH)2 =  0.080 L x 5.00^-3 M

                              = 0.0004 moles

From the reaction; 2 moles of HNO3 reacts with 1 mole of Ca(OH)2

Thus; moles of HNO3 = 0.0004 Moles × 2

                                    = 0.0008 moles

Volume = moles/ molarity

Therefore; volume of HNO3= 0.0008 Moles/ 5 × 10^-3 M

                                              = 0.16 L or 160 mL

what are the two models

explanation: