Be sure to answer all parts. find the ph of the equivalence point and the volume (ml) of 0.0895 m koh needed to reach the equivalence point in the titration of 23.4 ml of 0.0390 m hno,. volume: ml koh ph =

pH = 7

Volume of 0.0895 M KOH = 2.15 mL

Explanation:

(Assuming HNO, should be HNO₃)

The balanced chemical equation is:

HNO₃ + KOH ⇒ H₂O + KNO₃

At the equivalence point, the amount of KOH added is equivalent to the amount of HNO₃ present, which is calculated as follows:

(23.4mL)(0.0390mol/L) = 0.9126 mmol HNO₃

Since HNO₃ and KOH react in a 1:1 molar proportion, 0.9126 mmol of KOH is required to reach the equivalence point. Therefore, the volume of KOH solution required is:

(0.1926mmol) ÷ (0.0895mol/L) = 2.15 mL KOH solution

The HNO₃ and KOH react completely and neither is in excess, so only the products of the reaction remain (H₂O and KNO₃). Since KNO₃ can be considered a neutral salt, the pH at the equivalence point is 7.

= 49.674 g NaCl

Explanation:

From the equation;

1 mole of Sodium metal produces 2 moles sodium chloride

This means;

23 g of Na will produce 116.88 g of NaCl

Therefore;

11.5 g will generate;

 = (11.5 × 116.88)/23

 =  58.44 g of NaCl

But;

Percentage yield = (Actual yield/Theoretical yield)× 100%

         85 /100 = Actual yield /58.44 g

Thus;

Actual yield = 0.85 × 58.44

                  = 49.674 g NaCl