Calculate the pH of a titration at the point when 15.0 mL of 0.15 M NaOH is added to 30.0 mL 0f 0.20 M HNO.

The correct answer is 1.10.

Explanation:

Based on the given information, the molarity of the NaOH is 0.15 M, that is, 0.15 moles per liter of the solution.

Now the moles present in the 15 ml of the solution will be,  

0.015 × 0.15 = 2.25 × 10⁻³ moles of NaOH or 0.0025 moles of NaOH

Now, molarity of the HNO₃ given is 0.20 M, which means 0.2 moles per liter of the solution.  

Now the moles present in the 30 ml of the solution will be,  

0.030 × 0.2 = 0.006 moles of HNO₃

Now the complete disintegration of acid and base will be,  

NaOH (aq) (0.025 moles) ⇔ Na⁺ (aq) (0.025) + OH⁻ (aq) (0.025 moles)

HNO₃ (aq) (0.006 moles) ⇔ H⁺ (0.006 moles) + NO₃⁻ (aq) (0.006 moles)

Now the additional Hydrogen ions at titration point is,  

= 0.006 - 0.0025 = 0.0035 moles of H+

Now the concentration of H+ ions in the 45 ml of the solution will be,  

= 0.0035/45 × 1000

= 0.078 M

pH = -log[H⁺] = -log [0.078]

= 1.10

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.