Question

Use the References to access important values if needed for this question Determine the pH during the titration of 20.6 mL of 0.358 M hydroiodic acid by 0.702 M potassium hydroxide at the following points: (1) Before the addition of any potassium hydroxide (2) After the addition of 5.25 mL of potassium hydroxide (3) At the equivalence point (4) After adding 13.1 mL of potassium hydroxide

Answer 1

Soluhon is acidic, PHC7 M = pH = -loght] & pH = -log[w] V mL of M, alid u mixed with Vq mL of Ma base, the concentration of resultant solchon is given by Miv, -Mave if Ming > M₂ V₂ & S th M = " - May ₂ - Min if M₂ Y > MzY & Solchon is basic pt> 7. Given that [HI] = 0.358 a (H+] = 0-3584 Initial pH of HI - pt = -log 0.358 = 0.45 th 0.858 -5.25*0.702 -0.143 M. 2) M = 2016 20.6 +5.25 M = 20.680.358 = 0.054. 20.6+ 5.25 Sina on, u, M₂ , the solution is acidic pH = -log (tt) = ~ 10g0.143 0.845. 13.1X0.702 - 20.6x0.358 M= - 13.1 +20.6 Since M₂ V₂ & MV, the Solnhon 5 banc. poh - Lag (ort) = -logo.o54 & 1.27 pH - 14-POH - 14-1.27 & 12. I

HI --------> H⁺ + I^{​​​​​​-}

KOH -----------> K⁺ + OH^{​​​​​-}

Neutralization Reaction is

HI + KOH --------> KI + H₂O

3)

At equivalence point,

Acid is completely neutralized with base.

Hence [H⁺] = [OH^{​​​​​-}]

Hence pH = 7