Question

QUESTION 6 What is the pH of a buffer that has 0.250 mol HF and 0.100 mol NaF in 1.00 L that has 0.020 mol NaOH added to it? Ka=6.4x10-*

Answer 1

in UL doo day here on 0.250 mol HF 0.100 mor Mat C ya 0.020 moi Naolt ka= 6.4816 We know HF + NaOH Caw Na feast how Coy cap I mole of the reacts with Imole of NagH to produce i mole of Naole. me know Canv - nac.v "HE = 0.250.TL = 0.250 moles of HF na = 0.250 1l = 0.250 moves or F now we are addig 0.020 moles or naolt to the burren as bewen moks or strong base is there than weak acia, Naot will be completely consumed THE will be charged to THE= 0.375-0.0.250-0.020 = 0.23 mole, no.ox moles on tinoruds will increase ME = 0.250 +0.020 = 0.27 molas

new molan ities weak aud could be: . or and conjugate bain CHF ] - 0.23 (f) e 0.27 0.23 m 0.27 m Henderson - Hasselbalch equation pH a Pkatlog (conjugate based reak acid] -hop (6.") (627) le E-log fry 3.20 +0.07 = 3.27 pH = 3.27 ( Answers my answer Please kindly a Thumbs give up