A solution was prepared by dissolving 1210 mg of K_{3}Fe(CN)_{6} (329.2 g/mol) in sufficient waterto give 775 mL. Calculate: a) the molar analytical c

asked 2020-12-12
A solution was prepared by dissolving 1210 mg of \(\displaystyle{K}_{{{3}}}{F}{e}{\left({C}{N}\right)}_{{{6}}}\) (329.2 g/mol) in sufficient waterto give 775 mL. Calculate:
a) the molar analytical concentration of \(\displaystyle{K}_{{{3}}}{F}{e}{\left({C}{N}\right)}_{{{6}}}\).
b) the molar concentration of \(\displaystyle{K}^{{+}}\).
c) the molar concentration of \(\displaystyle{F}{e}{{\left({C}{N}\right)}_{{6}}^{{{3}-}}}\).
d) the weight/volume percentage of \(\displaystyle{K}_{{{3}}}{F}{e}{\left({C}{N}\right)}_{{{6}}}\).
e) the number of millimoles of \(\displaystyle{K}^{{+}}\) in 50.0 mL of thissolution
f) ppm \(\displaystyle{F}{e}{{\left({C}{N}\right)}_{{6}}^{{{3}-}}}\).
g) pK for the solution
h) \(\displaystyle{p}{F}{e}{\left({C}{N}\right)}_{{{6}}}\) for the solution.

Expert Answers (2)


I am using ur calculated value
Molarity of \(\displaystyle{F}{e}{{\left({C}{N}\right)}_{{{6}}}^{{{3}-}}}={4.74}\cdot{10}^{{-{3}}}\) M
Moles = molarity * V(L)
\(\displaystyle={4.74}\cdot{10}^{{-{3}}}{M}\cdot{0.775}{L}={3.674}\cdot{10}^{{-{3}}}\) mol
Mass of \(\displaystyle{F}{e}{{\left({C}{N}\right)}_{{{6}}}^{{{3}-}}}\) = moles * molar mass of \(\displaystyle{F}{e}{{\left({C}{N}\right)}_{{{6}}}^{{{3}-}}}\)
\(\displaystyle={3.674}\cdot{10}^{{-{3}}}\) mol * 212.0 g/mol
= 0.7788g
ppm = mass of solute / mass of solution \(*10^{6}\)
= 0.7788g / 775 g \(\displaystyle\cdot{10}^{{{6}}}\) (for water 1ml = 1g)
= 1005 ppm
= 1.005 mg/ L
Hope its clear to U!
for the molarity of \(\displaystyle{F}{e}{{\left({C}{N}\right)}_{{{6}}}^{{{3}-}}}\). , i got \(\displaystyle{4.74}{x}{10}^{{-{3}}}\) M

Best answer


Given values:

Mass of \(K_3Fe(CN)_6 = 1210 mg = 1.210 g  \)             (Conversion factor: 1 g = 1000 mg)

Molar mass of \(K_3Fe(CN)_6 = 329.2 g/mol\)

Volume of solution = 775 mL

Plugging values in equation 1:


\(Molarity \ of \ K_3Fe(CN)_6=\frac{1.210 \times 1000}{329.2 \times 775}\)


\(Molarity \  of \ K_3Fe(CN)_6=0.0047 M\)


The equation for the dissociation of \(K_3Fe(CN)_6\) into its ions follows:

\(K_3Fe(CN)_6 (aq) \rightarrow 3K^+ (aq) + Fe(CN)_6^{3-}(aq)\)

1 mole of \(K_3Fe(CN)_6\) produces 3 moles of K^+ ions and 1 mole of \( Fe(CN)_6^{3-}\) ion


Molar concentration of \(K^+  \ ions = (3\times 0.0047) = 0.0141 M\)



Molar concentration of \(Fe(CN)_6^{3−} \ ions = (1 \times 0.0047) = 0.0047 M\)


expert advice

Have a similar question?
We can deal with it in 3 hours

Relevant Questions

asked 2021-02-14
Mg reacts with H+ (aq) according to
Suppose that 0.524 g of Mg is reacted with 60.o ml of 1.0 M H+(aq). Assume that the density of the H+ (aq) solution os 1.00 g/ml,and that its specific heat capacity equals that of water. Theinitial and final temperatures are 22.0 degree celsius and 65.8degree celsius.
a) Is the reaction endothermic or exothermic?
b) Calculate \(\displaystyle\triangle{H}\) of of th reaction. Use correct sigs andgive units.
c) Calculate the \(\displaystyle\triangle{H}\) of the reaction per mole of magnesium.
asked 2021-05-16
Consider the curves in the first quadrant that have equationsy=Aexp(7x), where A is a positive constant. Different valuesof A give different curves. The curves form a family,F. Let P=(6,6). Let C be the number of the family Fthat goes through P.
A. Let y=f(x) be the equation of C. Find f(x).
B. Find the slope at P of the tangent to C.
C. A curve D is a perpendicular to C at P. What is the slope of thetangent to D at the point P?
D. Give a formula g(y) for the slope at (x,y) of the member of Fthat goes through (x,y). The formula should not involve A orx.
E. A curve which at each of its points is perpendicular to themember of the family F that goes through that point is called anorthogonal trajectory of F. Each orthogonal trajectory to Fsatisfies the differential equation dy/dx = -1/g(y), where g(y) isthe answer to part D.
Find a function of h(y) such that x=h(y) is the equation of theorthogonal trajectory to F that passes through the point P.
asked 2021-09-26
Calculate the molality of a solution that 1.35 M and has a density of 1.28 g/mL. It is also known that the molar mass of the solute is 100.0 g/mole
asked 2021-09-21
Analysis of a compound used in cosmetics reveals the compound contains 26.76% C, 2.21% H, 71.17% O and has a molar mass of 90.04 g/mol. Determine the molecular formula for this substance.