When pure HA is added to the buffer, the buffer component ratio and the pH decrease.
<h3>State and explain the relative change in the pH and in the buffer-component concentration ratio, [NaA]/[HA] for the dissolve of pure HA in the buffer.</h3>
When pure HA is added to the buffer, the buffer component ratio and the pH decrease. The added HA increases the concentrations of NA and HA. However, there is a greater relative increase in the concentration of HA. Hence, the ratio of [NaA]/[HA] decreases, causing the solution to become more acidic.
The capacity of a buffer to withstand pH change is measured. The concentration of the buffer's components namely, the acid and its conjugate base determine this ability. Greater buffer capacity is associated with higher buffer concentration.
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Answer:
C. If Assertion is true statement but Reason is false.
Explanation:
The burning of magnesium in air produces magnesium oxide as shown by the equation;
2Mg(s) + O2(g) -----> 2MgO(s)
The magnesium oxide solid is recovered as a white ash. This is a typical example of an oxidation reaction which is also a combustion reaction.
The reason has nothing at all to do with the assertion hence the answer given.
Magnesium oxide is basic just like the oxides of other metals and dissolves in water to yield an alkali.
Answer:
The correct answer is - may not be typical, and participant burden.
Explanation:
The 24-hour recall is nothing but a retrospective method of diet assessment. In this method, an individual is interviewed about his or her diet consumption during the last 24 hours.
The disadvantages or limitations of this method include the inability of a single day's intake to describe the typical diet, multiple recalls to intake, cost and administration time; participant burden, have to recall to reliably estimate usual intake.
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
