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yuradex [85]
2 years ago
5

A chemist prepares a solution of sodium nitrate (NaNO3) by measuring out 286. umol of sodium nitrate into a 450. mL volumetric f

lask and filling the flask to the mark with water.Calculate the concentration in mmol/L of the chemist's sodium nitrate solution.
Chemistry
1 answer:
shusha [124]2 years ago
3 0

Answer:

Concentration of sodium nitrate solution is 0.636 mmol/L

Explanation:

We know, 1micromol=1\times 10^{-3}mmol

So, 286.micromol=(286.\times 10^{-3})mmol=0.286mmol

Concentration in mmol/L is defined as number of mmol of solute dissolved in 1000 mL of solution

Here solute is NaNO_{3}

Total volume of solution = Total volume of volumetric flask = 450. mL

Hence concentration of NaNO_{3} solution = \frac{0.286mmol}{450.mL}\times 1000=0.636mmol/L

So, concentration of sodium nitrate solution is 0.636 mmol/L

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<h3>What is ecology and what are some terms in ecology?</h3>

Ecology is the study of living things and their interactions with their environment.

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Learn more about Peak phenomena, Biological surplus, Climax community and Sigmoid phenomena at: brainly.com/question/7149100

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How many atoms are in 15.0 moles of C2H6O
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determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
Mariana [72]

Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

              HNO₂ ⇄    H⁺   +   KNO₂⁻

C(i)        0.55M       0M      0.75M

ΔC            -x            +x          +x

C(eq)  0.55M - x       x     0.75M + x    b/c [HNO₂] / Ka > 100, the x can be                                    

                                                             dropped giving ...

           ≅0.55M        x       ≅0.75M        

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

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2 years ago
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