Question

Please help!

Answer 1

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 =$\frac{0.177 g}{174.24 g/mol}=0.001016 mol$

b) 1 mol = 1000 milli moles

0.001016  mol =0.001016 × 1000 milli moles = 1.016 milli moles

c) $Molarity=\frac{n}{V}$

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)

$[K_2SO_4]=\frac{0.001016 mol}{0.775 L}=0.001311 mol/L$

$K_2SO_4\rightarrow 2K++SO_4^{2-}$

1 mol potassium sulfate gives 2 mol of potassium ions and 1 mole of sulfate ions.

Molarity of potassium ions:

$[K^+]=2\times [K_2SO_4]=2\times 0.001311 mol/L=0.002622 mol/L$

Molarity of sulfate ions:

$[SO_4^{2-}]=1\times [K_2SO_4]=1\times 0.001311 mol/L=0.001311 mol/L$

d) $ppm=\frac{\text{Amount of solute in mg}}{\text{Volume of the solution in L}}$

Volume of solution = 0.775 L

Amount of potassium sulfate = 177 mg

$ppm=\frac{177 mg}{0.775 L}=228.39 mg/L$

e) Volume of the solution = 775 mL

Mass of potassium sulfate = 177 mg =0.177 g

$(w/w)\%=\frac{\text{Mass of the solute}}{\text{Volume of the solution in mL}}\times 100$

$(w/w)\%=\frac{0.177 g}{775 mL}\times 100=0.023\%$

f) $pK^=-\log[K^+]$

$pK^+=-\log[0.002622 M]=2.58$

$pSO_4^{2-}=-\log[SO_4^{2-}]$

$pK^+=-\log[0.001311 M]=2.88$

Answer 2

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