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please helpppp chemmmm?
1. How many moles of NaOH must be added to 0.400 L of 0.189 mol/L HF(aq) to obtain a solution with pH=3.4? Assume the temperature is 25oC. (For HF, pKa=3.18 at 25oC).
2. What is the pH at the equivalence point when when 40.0 mL of 0.189 mol/L HC3H5O2(aq) is titrated with 0.121 mol/L NaOH(aq)? Assume the temperature is 25oC. (For HC3H5O2, pKa=4.89 at 25oC).
3. If 75.0 mL of 0.117 mol/L NaOH(aq) and 45.0 mL of 0.183 mol/L HA(aq) are mixed, then what is [HA] at equilibrium? Assume that HA is a weak monoprotic acid with pKa=6.01 at 25oC.
can anyone answer the third one please?
1 Answer
- Dr.ALv 71 decade agoFavorite Answer
3.4 = 3.18 + log [F-]/ [HF
3.4 - 3.18 = 0.22 = log x/ 0.189 -x
10^0.22 =1.66 = x / 0.189-x
0.314 - 1.66x = x
x = [F-]= 0.118 M
moles F- = moles NaOH that must be added = 0.118 M x 0.400 L= 0.0472
mass NaOH = 0.0472 x 40 g/mol= 1.89 g
moles acid = 0.0400 L x 0.189 M=0.00756
moles NaOH required = 0.00756
Volume NaOH = 0.00756 / 0.121 M=0.0625 L
total volume = 0.0625 + 0.0400 =0.103 L
[C3H5O2-]= 0.00756/ 0.103 =0.0734 M
Ka = 10^-4.89=1.29 x 10^-5
Kb of C3H5O2- = Kw/Ka = 7.76 x 10^-10
C3H5O2- + H2O <=> HC3H5O2 + OH-
7.76 x 10^-10 =x^2/ 0.0734-x
x = [OH-]= 7.55 x 10^-6 M
pOH = 5.12
pH = 14 - pOH = 8.88
hope helps