Answer:
Conditioning two or three times will insure that the concentration of titrant is not changed by a stray drop of water.
Explanation:
"Check the tip of the buret for an air bubble. To remove an air bubble, whack the side of the buret tip while solution is flowing".
Answer:
more/less depending on if it lost or gained electrons during the ionic bond process
Explanation:
The solution of the lactic acd and sodium lactate is referred to as a buffer solution.
A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. In this case, the weak acid is the lactic acid and the conjugate base is the sodium lactate.
Buffer solutions are generally known to resist change in pH values.
When a strong base (in this case, NaOH) is added to the buffer, the lactic acid will give up its H+ in order to transform the base (OH-) into water (H2O) and the conjugate base, so we have:
HA + OH- → A- + H2O.
Since the added OH- is consumed by this reaction, the pH will change only slightly.
The NaOH reacts with the weak acid present in the buffer sollution.
Friction because smooth ice provides very little resistance against objects, like ice skates being dragged across its surface.
Answer:
13.20
Explanation:
Step 1: Calculate the moles of Ba(OH)₂
The molar mass of Ba(OH)₂ is 171.34 g/mol.
0.797 g × 1 mol/171.34 g = 4.65 × 10⁻³ mol
Step 2: Calculate the molar concentration of Ba(OH)₂
Molarity is equal to the moles of solute divided by the liters of solution.
[Ba(OH)₂] = 4.65 × 10⁻³ mol/60 × 10⁻³ L = 0.078 M
Step 3: Calculate [OH⁻]
Ba(OH)₂ is a strong base according to the following equation.
Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻
The concentration of OH⁻ is 2/1 × 0.078 M = 0.16 M
Step 4: Calculate the pOH
pOH = -log OH⁻ = -log 0.16 = 0.80
Step 5: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - 0.80 = 13.20
