Respuesta :
a. 599 mg x 1g/1000mg x 1mole/174.24g = 0.003438 moles
b. 0.003438 moles x 1000 mmoles/mole = 3.438 moles
c. 0.003438 moles/0.741 L = 0.0046 M
d. [K+] = 2 x 0.0046 = 0.0092 M because each mole K2 SO4 yields 2 moles K+
e. [SO42-] = 0.0046 M
f. 599 mg/0.741 L = 808.4 mg/L = 808 ppm
g. 0.599g/741 ml x 100 ml = 0.08 g/100 ml = 0.08% w/v
h. pK+ = -log [K+] = -log 0.0092 = 2.04
i. pSO42- = -log [SO42-] = -log 0.0046 = 2.34
b. 0.003438 moles x 1000 mmoles/mole = 3.438 moles
c. 0.003438 moles/0.741 L = 0.0046 M
d. [K+] = 2 x 0.0046 = 0.0092 M because each mole K2 SO4 yields 2 moles K+
e. [SO42-] = 0.0046 M
f. 599 mg/0.741 L = 808.4 mg/L = 808 ppm
g. 0.599g/741 ml x 100 ml = 0.08 g/100 ml = 0.08% w/v
h. pK+ = -log [K+] = -log 0.0092 = 2.04
i. pSO42- = -log [SO42-] = -log 0.0046 = 2.34
Explanation:
a) Mass of potassium sulfate = 177 mg =0.177 g
1 g= 1000 mg
Molecular mass of potassium sulfate = 174.24 g/mol
Moles of potassium sulfate,n =[tex]\frac{0.177 g}{174.24 g/mol}=0.001016 mol[/tex]
b) 1 mol = 1000 milli moles
0.001016 mol =0.001016 × 1000 milli moles = 1.016 milli moles
c) [tex]Molarity=\frac{n}{V}[/tex]
n = moles of substance
V = Volume of the solution in L
Molarity of of potassium sulfate:
Volume of the solution = 775 mL = 0.775 L (Mass of solute is very less)
[tex][K_2SO_4]=\frac{0.001016 mol}{0.775 L}=0.001311 mol/L[/tex]
[tex]K_2SO_4\rightarrow 2K++SO_4^{2-}[/tex]
1 mol potassium sulfate gives 2 mol of potassium ions and 1 mole of sulfate ions.
Molarity of potassium ions:
[tex][K^+]=2\times [K_2SO_4]=2\times 0.001311 mol/L=0.002622 mol/L[/tex]
Molarity of sulfate ions:
[tex][SO_4^{2-}]=1\times [K_2SO_4]=1\times 0.001311 mol/L=0.001311 mol/L[/tex]
d) [tex]ppm=\frac{\text{Amount of solute in mg}}{\text{Volume of the solution in L}}[/tex]
Volume of solution = 0.775 L
Amount of potassium sulfate = 177 mg
[tex]ppm=\frac{177 mg}{0.775 L}=228.39 mg/L[/tex]
e) Volume of the solution = 775 mL
Mass of potassium sulfate = 177 mg =0.177 g
[tex](w/w)\%=\frac{\text{Mass of the solute}}{\text{Volume of the solution in mL}}\times 100[/tex]
[tex](w/w)\%=\frac{0.177 g}{775 mL}\times 100=0.023\%[/tex]
f) [tex]pK^=-\log[K^+][/tex]
[tex]pK^+=-\log[0.002622 M]=2.58[/tex]
[tex]pSO_4^{2-}=-\log[SO_4^{2-}][/tex]
[tex]pK^+=-\log[0.001311 M]=2.88[/tex]
