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kaheart [24]
2 years ago
6

What is the concentration in milligrams per milliter of a solution containing 23.5 meq sodium chlorise per milliliter? mw nacl =

58.5? 2
Chemistry
1 answer:
ValentinkaMS [17]2 years ago
7 0

1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.

It is calculated in mg/ml.

The unit of measurement frequently used for electrolytes is the milliequivalent (mEq). This value compares an element's chemical activity, or combining power, to that of 1 mg of hydrogen.

Formula for calculating concentration in mg/ml is

Conc. (mg/ml) = M(eq) /ml ×  Molecular weight / Valency

Given

M(eq) NaCl/ ml = 23.5

Molecular weight pf NaCl = 58.5 g/mol

Valency = 1

Putting the values into the formula

Conc. (mg/ml) = 23.5 ×58.5/1

                       = 1374.75 mg/ml

Hence, 1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Learn more about Concentration here brainly.com/question/14500335

#SPJ4

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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

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Answer:

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