Strong acid-titrated with strong base. suppose the titration was reversed in question 2. if you titrated 30.0 ml of 0.1 m hcl with 0.1 m naoh, indicate the approximate ph (a) at the start of the titration and (b) at the equivalence point. (c) what is the total volume of solution at the equivalence point? add this curve to your sketch in question 2. 3. la. 1.0; b. 7.0; c. 60 ml]

(a) pH = 1

(b) pH = 7

(c) 60 mL

Explanation:

(a) At the start of the titration, there is only 0.1 M HCl, which is a strong acid, according to the following expression.

HCl → H⁺ + Cl⁻

The concentration of H⁺ is 0.1 M as well and the pH is:

pH = -log [H⁺] = -log 0.1 = 1

(b) The reaction between HCl and NaOH is:

HCl + NaOH → NaCl + H₂O

At the equivalence point, all the HCl and NaOH have reacted. Since NaCl does not contribute to the pH, the pH is 7.

(c) The moles of HCl are:

(0.1 mol/L) × (30.0 × 10⁻³ L) = 3 × 10⁻³ mol

The molar ratio of HCl to NaOH is 1:1, so we also have 3 × 10⁻³ moles of NaOH. The volume of NaOH that contains these moles is:

3 × 10⁻³ mol / (0.1 mol/L) = 0.03 L = 30 mL

The total volume is:

V(HCl) + V(NaOH) = 30 mL + 30 mL = 60 mL

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.