Question

2) How many grams of Nat would have to be added to 2.00 Lof 0.100 MHF to yield a solution with a phi 4.00? Assume the volume of the solution remains at 2.00 Lufter the Naf is added. (a) 300 (b) 368 (c) 0.84 (d) 6.98 (e) 60. G 18. A titration of 20.00 mL of 0.075 M HNO2 (nitrous acid) with 0.0250 M N OH is performed. Given that Ka for nitrous acid is 4.5 x 10-4, the pH at 30.00 mL of NaOH added is (a) 1.94 (b) 7.00 (c) 3.35 (d) 4.50 le) 2.44

Answer 1

Ans2. Option e is correct option

Ans- [HF) = 0.100M Volume 2.00 L HE us weak acid and Nat us salt of strong base and weak acid. pH = 4.00 jpka of HE= 3.14 [salt) PH= pka + loq acid) PH = 3:14 + dog [Naf) [HE] leal [NAF) [0.100) [NaF ] 4.00 -3.14 = log log to 100 100:86 of mass [NAF) =) [NaF ) - 10786x 0.100 [0.100) [NaF) = 0.7244 M = 0.9244 mal Liten males [NqF] = 0-7244 mal x 2.00 Liten ūtn 31.4488 mal of [NaF] = males of wat males of Nat & malar mass ress of NF 1.4488 mal x 41.88817gmat' 60.8 = 60 bog mass of NaF is bog Thenefone (e) (609) us correct option. Ans.18. milimels of Ang = 20mL X 0.075M = 15 mibmaks milimales NaOH = 30 mx 0.0250 M = 0.75 milimales ka=4.5810 4 pkę = -dog (4-5x164) = 3-35 > HNO + NaOH Ž NaNO2 + H2O I 0.75 O -0.75 +0.75 -0.75 E 0.75 0.75 males of NaNO2 At equilibrium = 0.75 milimales moles of HNOL = 0'75 milimales- Tatal volume of solution = (20+30) mL = 50mL pH = pkat + dog platilag Salt ] acid -0.75 50 075 50 PH = 3.35 + • lag! pH = 3-35 to pH = 3-35 Therefore (C) 3:35 Jo correct oftion. Scanned with CamScanner CS Scanned with CamScanner

Ans. 18. Option C is correct Option

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