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Irina-Kira [14]
3 years ago
14

A solution that is 0.20 m in hcho2 and 0.15 m in nacho2 find ph

Chemistry
1 answer:
Mashutka [201]3 years ago
5 0
We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH

The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])

We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87

So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.
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3 years ago
Urgent please help me!!!
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How many grams in 82 moles of CO2
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What is the effect of on the volume of a<br> gas?
Scrat [10]

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3 years ago
Read 2 more answers
If a solution contains 3 moles/liter of sodium chloride (nacl, made of sodium ions and chloride ions), what is the osmolarity of
valkas [14]

<u>Answer:</u> Osmolarity of the sodium chloride solution is 18 Osmol/L.

<u>Explanation:</u>

Osmolarity is defined as the the concentration of the solution which is expressed as the total number of solution particles present in one liter of solvent.

We are given the molarity of the solution which is 3mol/L and to convert it into osmolarity, we will multiply the number of osmoles that are produced by the solute.

Osmole is defined as the particles that contribute to the osmotic pressure of a solution.

The solute given here is sodium chloride (NaCl). Number of osmoles can be determined by the dissociation of this solvent into ions.

The equation given by the dissociation of NaCl:

NaCl\rightarrow Na^++Cl^-

1 mole of sodium chloride produces 2 moles of ions.

So, 3 moles of sodium chloride will produce = (3 × 2) = 6 moles of ions.

Hence, osmolarity of the sodium chloride solution will be 6\times 3mol/L=18Osmol/L

5 0
3 years ago
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