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charle [14.2K]
2 years ago
6

Calculate the pH of a solution that has a [OH−] of 2.6 × 10^−6 M.

Chemistry
1 answer:
TiliK225 [7]2 years ago
7 0

Answer:

A. 8.4

Explanation:

[OH⁻] = 2.6 × 10⁻⁶               Take the negative log of each side

-log[OH⁻] = pOH = 5.59     Apply the pH/pOH relation

pH + pOH = 14.00               Insert the value of pOH

pH + 5.59 = 14.00               Subtract 5.59 from each side

pH = 14.00 -5.59 = 8.41

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You have been working to develop a new fictitious compound in the lab. Determine the amount in units of moles​ [mol] of 20 grams
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From the chemical formula the total mass of the compound can be determined. The mass of the 1 mole of the compound is its molar mass. The atom by which the molecule is generated, the mass of these atoms are expressed in terms of amu or atomic unit mass, but after formation of a molecule in a particular ratio the mass of each of the atom becomes the total molecular weight of the generated molecule. In this case the molecule posses three atoms X, Y and Z which are in a ratio of 2:2:7. Thus the chemical formula of the compound can be written as X_{2}Y_{2}Z_{7}.

So the total mass of the compound in amu is {(2×47)+(2×42)+(7×16)} = {94+84+112}=290 amu.

Thus 1 mole of the compound contains 290 amu or 290 g  by mass.

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Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in
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<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion

1) Content of Ca (2+) ions

Calcium chloride = CaCl2

Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)

=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)

Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution

M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2

=> 0.0825 mol Ca(2+)

2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)

formula of phospahte ion: PO4 (3-)

molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2

Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)

=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)

3) Content of Mg(2+) ions

Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)

Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)

number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution

n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2

ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)

4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)

2PO4(3-) + 3Mg(2+) = Mg3(PO4)2

=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)

=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)

5) Total number of moles of PO4(3-)

0.055 mol + 0.16 mol = 0.215 mol

6) Sodium phosphate

Sodium phosphate = Na3(PO4)

Na3PO4 ---> 3Na(+) + PO4(3-)

=> 1 mol Na3PO4 : 1 mol PO4(3-)

=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4

mass in grams = number of moles * molar mass

molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol

=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g

Answer: 35.26 g of sodium phosphate
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