1) A 25.0 mL sample of 0.150 M hydrocyanic acid is tritated with a 0.150 M NaOH

Question

1) A 25.0 mL sample of 0.150 M hydrocyanic acid is tritated with a 0.150 M NaOH solution. What is the pH before any base is added? The Ka of hydrocyanic acid = 4.9×10^-10

2) A 25.0 mL sample of 0.150 M hydrozoic acid is tritated with a 0.150 M NaOH solution. What is the pH after 26.0 mL of base is added? The Ka of hydrozoic acid = 1.9×10^-5

3) How many milliliters of 0.120 M NaOH are required to tritrate 50.0 mL of 0.0998 M hydrochloric acid to the equivale point the Ka of hypochlorous acid is 3.0×10^-8. What is the pH after?

Explanation / Answer

1) Total volume = 25 ml =0.025 L

Moles of H+ = 0.150 * 0.025 = 3.75*10-3 mol

Moles of CN- = 0.150*0.025 = 3.75*10-3 mol

[CN-] = 3.75*10-3 mol / 0.025 L = 0.150 M

pH = pka+log ([CN-]/[HCN])

= -log Ka + log ([CN-]/[HCN])

= -log 4.9*10-10 + log (0.150 M / 0.150M)

pH= 9.31

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