a) for weak acid ,
general solution :'
[H+] = ([HOBr]*Ka)^1/2 = (0.0168*2.1*10^-9)^1/2 = 5.94*10^-6 M
pH = -log [H+]
pH = 5.23
b)
moles of NaOH added :0.0100 moles
moles of HOBr in 1 L = 0.0168 moles
reaction : HOBr(aq) + NaOH (aq) ===> NaOBr (aq) + H2O (l)
now , the system behaves as buffer solution :
pH = pKa + log [NaOBr]/[HOBr]
pKa = -log Ka = 8.68
[NaOBr] = 0.0100 M
[HOBr] = 0.0168 - 0.0100 = 0.0068 M
pH = 8.68 + log(0.010/0.0068 )
pH = 8.85
urgent 3) Hypobromous acid (HOBr) is a weak monoprotic acid, with K. = 2.1 x 10°...
3) Hypobromous acid (HOBr) is a weak monoprotic acid, with K. = 2.1 x 10 a) What is the value for pH for a 0.0168 M aqueous solution of hypobromous acid? (12 points] b) 0.0100 moles of NaOH, a strong soluble base, is added to 1.000 L of the above solution of hypobromous acid. What will be the pH for this new solution? [15 points]
urgent
3) Hypobromous acid (HOBr) is a weak monoprotic acid, with K = 2.1 x 10-9 a) What is the value for pH for a 0.0424 M aqueous solution of hypobromous acid? (12 points) b) 0.0100 moles of NaOH, a strong soluble base, is added to 1.000 L of the above solution of hypobromous acid. What will be the pH for this new solution? [15 points]
3) Hypobromous acid (HOBr) is a weak monoprotic acid, with K. = 2.1 x 10-9 a) What is the value for pH for a 0.0424 M aqueous solution of hypobromous acid? [12 points) I b) 0.0100 moles of NaOH, a strong soluble base, is added to 1.000 L of the above solution of hypobromous acid. What will be the pH for this new solution? [15 points]
3) Hypobromous acid (HOB) is a weak monoprotic acid, with K.-2.1 x 10%. a) What is the value for pH for a 0.0168 M aqueous solution of hypotromous acid? [12 points] b) 0.0100 moles of NaOH, a strong soluble base, is added to 1.000 L of the above solution of hypobromous acid. What will be the pH for this new solution? [15 points) 10
3) Hypochlorous acid (HOCI) is a weak monoprotic acid, with K. = 3.0 x 10-8 a) What is the value for pH for a 0.0264 M aqueous solution of hypochlorous acid? [12 points] b) 0.0100 moles of NaOH, a strong soluble base, is added to 1.000 L of the above solution of hypochlorous acid. What will be the pH for this new solution? [15 points)
3) Hypobromous acid (HOBr) is a weak monoprotic acid, with Ka = 2.1 x 10^-9 . a) What is the value for pH for a 0.0424 M aqueous solution of hypobromous acid? [12 points]
3) Hypoiodous acid (HOI) is a weak monoprotic acid, with Ka = 2.3 x 10-". a) What is the value for pH for a 0.0424 M aqueous solution of hypoiodous acid? [12 points) b) 0.0100 moles of NaOH, a strong soluble base, is added to 1.000 L of the above solution of hypoiodous acid. What will be the pH for this new solution? [15 points)
2) Thermochemical data are given below (at T = 25.0 °C) and may be useful in doing the following problem. Substance ΔΗ, (kJ/mol) Sº (J/mol.K) AGⓇ (kJ/mol) Mg? (aq) Mg(OH)(s) OH(aq) - 466.9 - 924.7 - 230.0 - 138.1 63.1 - 10.8 - 454.8 - 833.9 - 157.2 Using only the data above, find AGºrn, AS, and K for the reaction Mg(OH)2(s) 5 Mg? (aq) + 2 OH(aq) [18 points) 3) Hypobromous acid (HOBr) is a weak monoprotic acid, with...
Part 3. Problems 1) The molarity of naphthaline in a solution of naphthalene (CH. MW-128.8), a non-ionizing and nonvolatile solute, and benzene (C.HS, MW = 78.11), a volatile solvent, is [CHx] = 0.125 M. The density of the solution is D= 0.8781 g/cm! Find the following: a) The osmotic pressure exerted by the solution, at T = 25.0°C. Give your final answer in units of torr. [8 points] b) Xn, the mole fraction of naphthalene in the solution. [15 points)...
2. Calculate the pH expected for the following monoprotic substances. a. HCl (strong acid) 0.0100 M 0.100 M 0.0500 M 0.00100 M b. NaOH (strong base) 0.100 M 0.0500 M 0.0100 M Experiment 4 с. НС2Н3О2 (weak acid) (Ka 1.8 x 10-5) 0.100 M 0.0500 M 0.0100 M d. K2CO3 (weak base) (K, = 2.1 x 10-4) 0.100 M 0.0500 M 0.0100 M
2. Calculate the pH expected for the following monoprotic substances. a. HCl (strong acid) 0.0100 M 0.100...