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Lilit [14]
3 years ago
13

A 0.89% (w/v) sodium chloride solution is referred to as physiological saline solution because it has the sameconcentration of s

alts as human blood. What is the molarity of a physiological saline solution?
Chemistry
1 answer:
elixir [45]3 years ago
4 0

Answer: Molarity of a physiological saline solution is 0.15 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

n = moles of solute

V_s =volume of solution in ml

Given : 0.89 g of sodium chloride in 100 ml of solution.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.89g}{58.5g/mol}=0.015moles

Volume of solution = 100 ml

Molarity=\frac{0.015\times 1000}{100ml}=0.15M

Molarity of physiological saline solution is 0.15 M

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A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
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1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

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