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Alenkinab [10]
3 years ago
9

The recommended daily intake of potassium ( K ) is 4.725 g . The average raisin contains 3.513 mg K . Fill in the denominators o

f the fractions, d 1 and d 2 , to complete the conversion factors that can be used to convert between grams and milligrams, given that 1 g is equivalent to 1000 mg . 1 g d 1 mg or 1000 mg d 2 g
Chemistry
1 answer:
kondor19780726 [428]3 years ago
4 0

Explanation:

It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.

             \frac{1 g}{1000 mg} or \frac{1000 mg}{1 g}

So, when we have to convert grams into milligrams then we simply multiply the digit with 1000. And, if we have to convert a digit from milligrams to grams then we simply divide it by 1000.

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Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO2H) and 0.230 mol of sodium formate (NaCO2H)
Veronika [31]

The question is incomplete, here is the complete question:

Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO₂H) and 0.230 mol of sodium formate (NaCO₂H) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is 1.77 × 10⁻⁴

a) 2.099

b) 10.463

c) 3.546

d) 2.307

e) 3.952

<u>Answer:</u> The pH of the solution is 3.546

<u>Explanation:</u>

We are given:

Moles of formic acid = 0.370 moles

Moles of sodium formate = 0.230 moles

Volume of solution = 1 L

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:  

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=pK_a+\log(\frac{[HCOONa]}{[HCOOH]})

pK_a = negative logarithm of acid dissociation constant of formic acid = 3.75

[HCOONa]=\frac{0.230}{1}  

[HCOOH]=\frac{0.370}{1}

pH = ?  

Putting values in above equation, we get:  

pH=3.75+\log(\frac{0.23/1}{0.37/1})\\\\pH=3.54

Hence, the pH of the solution is 3.546

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