Question

# 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

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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.

2020-12-13

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

2021-10-08

(a):

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:

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

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

b)

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 $$K^+ \ ions = (3\times 0.0047) = 0.0141 M$$

c)

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