What volume of 5.00×10−3mhno3 is needed to titrate 20.00 ml of 5.00×10−3mca(oh)2 to the equivalence point?

The balanced equation for the given titration reaction can be written as

Step 1 : Find moles of Ca(OH)₂

Volume of Ca(OH)₂ = 20 mL

Concentration of Ca(OH)₂ = 5.00 x 10⁻³ M

Moles of Ca(OH)₂ =

Moles of Ca(OH)₂ = 1.00 x 10⁻⁴

Step 2 : Find moles of HNO3 using mole ratio from balanced equation

The mole ratio of HNO₃ and Ca(OH)₂ is 2 : 1

Let us use this as a conversion factor to find moles of HNO₃.

Moles of HNO₃ =

Moles of HNO₃ = 2.00 x 10⁻⁴

Step 3 : Find concentration of HNO3

Concentration of HNO3 can be calculated as

The concentration of HNO₃ is given as 5.00 x 10⁻³

Moles of HNO3 = 2.00 x 10⁻⁴

Let us plug in these values in above equation.

Volume of HNO₃ in mL =

Volume of HNO₃ needed for titration is 40 mL

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.