J. Calculate the [H-] in a solution that is 0.17 Min NaE and 0.25 Min HF. 7.2 x0
ID: 1008491 • Letter: J
Question
J. Calculate the [H-] in a solution that is 0.17 Min NaE and 0.25 Min HF. 7.2 x010-4) a) 7.2 x 10M b) 1.5M 1.1x 103 M 4.9x 10 M d) 0.20M 2. What will happen if a small amount ofhydrochloric acid is added to a0.1 Msolution ofHF? a) Thepercentionization ofHF will increase. b) Thepercentionization ofHF will decrease. c) Thepercentionization ofHF will remain unchanged. d) for HF will increase. e) for HF will decrease. 3. Which of the followingwill not produce a buffered solution? a) 100 mL of0.1 MNa^CO3 and 50 mL of0.1 MHC b) 100mL of0.1 MNaHCO3 and 25 mL of0.2 MHC c) 100 mL of0.1 MNa^CO3 and 75 mL of0.2 MHC d) 50 mL of0.2 MNa CO3 and 5 mL of1.0 MHO e) 100 mL of0.1 MNa2CO3 and 50 mL of0.1 MNaQH 4. A 100. mL sample of0.10 MHCl is mixed with 50. mL of0.12 MNHs. What is the 10-2) resulting pH7 (Ke for NH, = 1 .8 x a) 3.07 b) 10.93 c) 12.43 d) 1.40 ) 1.57 5. The following question refers to a 2 .0-liter buffered solution created from 0 82 MNH, (K= 1 . 8 x 10%) and 0.26 MNH4F. What is the pH of this solution? a) 9.26 b) 9.75 c) 4.25 d) 5.24 e) 8.76Explanation / Answer
pH = pKa + log(salt)/(acid)
pH = -log(7.2*10^-4) + log((0.17)/(0.25))
pH = 2.975
[H+] = 10^-2.975 = 1.1*10^-3
Answer = C
If you add HCl (strong electrolyte) then the HF dissociation decreses due to increases in H+ ion concentration
Answer: B
Answer : E
Since both Na2CO3 and NaOH are basic solutions.Buffer solution means weak acid with its salt of strong base or weak base and its salt of strong acid.
number of moles of HCl = 100*0.1 = 10/150 = 0.067 mol
number of moles of NH3 = 50 * 0.12 = 6/150 = 0.04 mol
remaining moles of HCl = 0.067 - 0.04 = 0.027
pH = -log(H+) = -log(0.027)
pH = 1.57
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