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Phoenix [80]
3 years ago
9

You need to produce a buffer solution that has a pH of 4.97. You already have a solution that contains 10. mmol (millimoles) of

acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution?
Chemistry
1 answer:
motikmotik3 years ago
5 0

Answer:

16.9 mmoles are needed to add to the solution

Explanation:

To solve this, we can apply the Henderson Hasselbach equation:

Acetic acid → CH₃COOH

CH₃COOH  +  H₂O  ⇄  CH₃COO⁻   +  H₃O⁺             Ka

That is the acid base equilibrium, to determine the amount:

pKa of acetic acid is 4.74

Henderson Hasselbach formula is:

pH = pKa + log (base/acid)

We replace  →  4.97 = 4.74 + log (B / 10mmoles)

Let's verify mmoles of B

4.97 - 4.74 = log (B/10)

0.23 =  log (B/10)

10^(0.23) = 10^(log (B/10)

1.698 = B/10

B = 1.698 . 10 → 16.9 mmoles

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4 0
2 years ago
What volume of 12 M NaOH and 2 M NaOH should be mixed to get 2 litres of 9 M NaOH solution?
Sergio [31]
8/5lit.. of 12M NaOH
2/5lit.. of 2M NaOH
7 0
3 years ago
What is true at stp? A)the temperature is 273.15 kelvin, B)one mole of gas occupies one liter of volume. C)pressure is at 101 vo
fgiga [73]
STP stands for standard temperature and pressure. Standard pressure is equivalent to 1 atm, and standard temperature is equivalent to 273.15 K. Therefore, your answer is A. the temperature is 273.15 kelvin.

Hope this helps!
3 0
3 years ago
Which of the following is a salt? <br> C6H12O6<br><br> O2<br><br> C2H2<br> K C l
irina [24]

Answer:

it's kcl

Explanation:

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5 0
2 years ago
Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in
avanturin [10]
<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
</span>
5 0
3 years ago
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