Will give BRAINLIEST A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water.
Answer the following questions, being sure to include units and remember sig figs. Show your work.

What possible issues may result from air bubbles in the burette for this experiment? Be specific about
how this may affect the final calculated concentration of acid (Will it be lower or higher than actual?
Why?)

The calculated concentration of acid will be higher than the actual concentration of acid

Explanation:

We have information that all enable us to calculate the concentration of KOH in the solution. From the question, we have;

Mass of KOH= 14.555g

Molar mass of KOH= 56.1056 g/mol

Volume of solution= 500 ml

Number of moles of KOH= ???

From;

m/M= CV

m= mass of KOH

M= molar mass of KOH

C= concentration of KOH solution

V= volume of solution

Substituting values;

14.555g/56.1056 g/mol = C× 500/1000

0.259 moles = 0.5C

C= 0.259/0.5

C= 0.518 M

If the acid is HA, the reaction equation is;

KOH(aq) + HA(aq) > KA(aq) + H2O(l)

The concentration of the acid is usually determined via titration. This involves delivering a particular volume of acid in a burette into the base and watching out for the volume of acid used at end point. If there are air bubbles in the burette, then more volume of acid is recorded than that actually used and this will make the calculated concentration of the acid to be higher than the actual concentration of acid present.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.