Consider a solution made by mixing HCN (Ka = 6.2 × 10–10) with
HC2H3O2 (Ka = 1.8 × 10–5) in aqueous solution. What are the major
species in solution? H+, CN–, H+, C2H3O2–, OH–, H2O
H+, CN–, HC2H3O2, H2O
HCN, HC2H3O2, H2O
H+, CN–, H+, C2H3O2–, H2O
HCN, H+, C2H3O2–, H2O
Please explain with your answer.
HCN has a pKa = 6.2 x 10-10. If a 50.0 mL
of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the
pH a) after 8.00 mL base, b) at the halfway point of
the titration, c) at the equivalence point.
Answers are below, I need to know how to get to them step by
step/conceptually
Confirm: a) 8.49 , b) 9.21 c) 10.96
Consider the reaction HCN(aq) H+(aq) +CN-(aq). the equilibrium
constant is 6.2*10^-10. If you place 0.4 mols of HCN in a 2.0 liter
flask, what is the equilbrium of CN-? so far ive got to
x=x(0.2)(6.2*10^-10)
1. Ka for hydrocyanic acid,
HCN, is
4.00×10-10. Ka
for acetylsalicylic acid (aspirin),
HC9H7O4, is
3.00×10-4. Ka for
phenol (a weak acid),
C6H5OH, is
1.00×10-10
What is the formula for the
strongest acid?
2. Ka for nitrous acid,
HNO2, is
4.50×10-4. Ka for
hydrocyanic acid, HCN, is
4.00×10-10. Ka for
phenol (a weak acid),
C6H5OH, is
1.00×10-10.
What is the formula for the
strongest conjugate base?
3. The compound ammonia ,
NH3, is a weak base when dissolved in
water. Write...
Consider the dissociation of aqueous HCN at 25°C:
HCN(aq) --> H+(aq) + CN–(aq) K =
6.2×10–10 .
a) Compute ΔG° at 25°C.
b) If 0.120 mol of HCN is dissolved to make 200. mL of solution,
then what are the equilibrium concentrations of all of the
species?
c) Suppose 0.040 mol of HCN is dissolved in the same solution
without appreciably increasing the volume. Compute the derivative
dG/dξ for the system before it has a chance to
re-establish equilibrium. Use...
consider the titration of a 20ml sample of .105M HCN with .125M
NaOH(ka=4.9*10^-10 of HCN)
Draw a scheme of how the titration curve should look like
Find:
initial pH, pH when 10ml of NaOH were added, pH at the
equivalence point, pH when 16.8ml of NaOH were added
A student is given 500.mL
of a 0.10M
HCN solution. Ka =
4.9 X 10^−10
What mass of NaCN
should the student dissolve in
the HCN solution to turn it into a buffer with pH =
9.86?
what is the ph of a 0.10m solution of nacn at 25°c
(ka=4.9×10^-10 for hcn). I've gotten 5.15 as a pH but the correct
answer is 11.15. I don't see how it's 11.15...