Lec 1: Laboratory Mathematics Flashcards

1
Q

How many grams of NaCl will be needed to prepare 100mL of a 10% (w/vol.) solution?

A

10% = (g of NaCl/100mL) x 100 10% x 100mL = 10 grams of NaCl

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2
Q

How many grams of NaOH will be needed to prepare 200mL of a 15% (w/vol.) solution

A

15% = (g of NaOH/200mL) x 100 15% x 200mL = 30 grams of NaOH

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3
Q

How much HCl will be needed to prepare 1L of a 0.25 solutionM?
○ MW=36
○ SG = 1.19
○ %Assay = 37%

Also the liquid form.

A

0.25M = (x)/(36g x 1L)
0.25M x (36g x 1L) = 9 grams of HCl
Further conversion: 9 grams/ 1.19 = 7.56 mL 7.56 mL/37% = 20.44mL of liquid HCl

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4
Q

How much CH3COOH will be needed to prepare 800mL of a 0.5M solution?
○ MW=60
○ SG = 1.06
○ %Assay = 99%

+ liquid form

A

0.5M = (x)/(60g x 0.8L)
0.5M x (60g x 0.8L) = 24 grams of
CH3COOH
Further conversion: 24 grams/1.06 = 22.64 mL
22.64mL/99% = 22.87mL of liquid CH3COOH

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5
Q

How much H2SO4 will be needed to prepare 1L of a 0.35N solution?
○ MW=98
○ SG = 1.84
○ %Assay = 96%

+ liquid form

A

0.35N = (x)/(49g x 1L)–sulfuric acid has 2 replaceable hydrogens
0.35M x (49g x 1L) = 17.15 grams of H2SO4

Further conversion: 17.15 grams/1.84 = 9.32mL 9.32mL/96% = 9.71mL of H2SO4

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6
Q

How much HNO3 will be needed to prepare 500mL of a 0.5N solution?
○ MW=63
○ SG = 1.42
○ %Assay = 70%

+ liquid form

A

0.5N = (x)/(63g x 0.5L)
0.5N x (63g x 0.5L) = 15.75 grams of
HNO3

Further conversion: 15.75 grams/1.42 = 11.09mL
11.09mL/70% = 15.84mL of HNO3

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7
Q

What will be the final volume if the 100mL 10% NaCl you prepared is to be reduced to 5%?

A

Using the C1V1 = C2V2 formula: (10%)(100mL) = (5%)(x)
x = (10%)(100mL)/5%
x = 200mL

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8
Q

Convert 0.35N of H2SO4 into M

A

Normality = Molarity x Valence M = 0.35/2
M = 0.175

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9
Q

Convert the prepared 10% NaCl into M

A

M = (%w/v x 10)/ MW M = (10% x 10)/58.44g
M = 0.0172

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10
Q

Convert 325 mg/dL of Na into mEq/L

A

mEq/L = (mg/dL x 10)/Eq. weight mEq/L = 325 x 10/23 (atomic mass of Na) mEq/L = 141.30

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11
Q

Convert 100 mg/dL of glucose into mmol/L (MW = 180)

A

mmol/L = (100 x 10)/180
mmol/L = 5.56

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12
Q

Convert 5 mEq/L of Ca into mmol/L

A

mEq/L = mmol/L x V

mmol/L = (5 mEq/L)/2 (Ca is divalent) mmol/L = 2.5

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13
Q

If 0.9mL of NSS is added to 0.1ml of serum, what is the dilution?

A

D = 0.1mL/(0.9mL + 0.1mL) D = 0.1mL/1mL
D = 1/10 or 0.1 (look for the LCF for the numerator and denominator)

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14
Q

How much serum is needed to prepare a 1⁄5 dilution with a volume of 2ml?

A

1/5 = (x)/(2mL) x = (1⁄5) x 2mL
x = 0.4mL

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15
Q

A 0.5mL serum sample with a glucose value of 100mg/dL is diluted with 0.5mL NSS (Tube 1). The sample from Tube 1 is subsequently diluted with 1mL of NSS Tube 2

  1. What is the initial dilution?
  2. What is the solution dilution in Tube 2?
  3. What is the final concentration of
    glucose?

What if there was a Tube 3 with the ff. specifications:
● Tube 1 still has 100mg/dL of glucose
● Tube 3 has taken 1mL from Tube 2 and
added 1mL of NSS

A

Answer to 1: D = 0.5mL/0.5mL + 0.5mL ■ D = 0.5mL/1mL
D = 1/2

Answer to 2: D = 0.5mL/1mL + 0.5mL D = 0.5mL/1.5mL
D = 1/3 x (1/2)–multiply to the previous dilution
D = 1/6

Answer to 3: 100mg/dL x 1⁄6 = 16.67mg/dL

Answer: D = 1mL/1mL + 1mL D = 1/2
D=1⁄2x1⁄3x1/2
D = 1/12 or 0.083

Final Concentration = 100mg/dL x 1/12 Final Concentration = 8.33mg/dL

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16
Q

Determine the pH, pOH, [H], and [OH] of 0.0035M HCl

A

Answer to 1:
● pH = -log(0.0035)
pH = 2.46
● pOH=14-2.46
pOH = 11.54
● [H] = 0.0035M
● [OH] = 10^-11.54
[OH] = 2.88 x 10^-12 (the negative exponent can be derived from finding the next whole number after the pOH/pH value)

17
Q

Determine the pH, pOH, [H], and [OH] of 0.00095M NaOH

A

Answer to 2:
● pH=14-3.02
pH = 10.98
● pOH = -log(0.00095) pOH = 3.02
● [H] = 1 x 10^-14/0.00095M [H] = 1.05 x 10^-11
● [OH] = 0.00095M

18
Q

Determine the pH, pOH, [H], and [OH] of 0.0075M HOAc with a Ka of 1.82 x 10^-5

A

Answer to 3:
● pH = -log(3.69 x 10^-4)
pH = 3.43
● pOH=14-3.43
pOH = 10.57
● [H] = sqrt. (1.82 x 10^-5)(0.0075M)
[H] = 3.69 x 10^-4
● [OH] = 10^-10.57
[OH] = 2.69 x 10^-11

19
Q
  1. What is the pH of a buffer composed of 0.38M HOAc and 0.40M NaOAc with a pKa value of 4.74?
  2. What will be the final pH if 10mL of 0.15M HCl is added to 100mL of the original buffer?
A

Answer to 1:
● pH = 4.74 + log(0.40M/0.38M)
pH = 4.76

Answer to 2:
● Determine the concentrations:
○ 0.15M of HCl x 10mL = 1.5mmol
○ 0.38M of HOAc x 100mL = 38 mmol
○ 0.40M of NaOAc x 100mL = 40 mmol
● Add and subtract accordingly to get the new concentrations:
○ HOAc = 38 + 1.5 = 39.5mmol
○ NaOAc = 40 - 1.5 = 38.5mmol
● Get the pH:
● 4.74 + log(38.5/39.5) = 4.73