Question

explain how and why the naoh molarity for part a would or would not be affected if 50 ml of water were added to dissolve the khp in part a instead of 30 ml.

Explanation:

1. For the first part, the molarity of NaOH would not be affected. KHP (potassium hydrogen phthalate) is a primary standard. The amount of KHP is determined by its mass, not the volume of water used to dissolve it. The reaction between KHP and NaOH is stoichiometrically - based on the moles of KHP present. Adding more water (from 30 mL to 50 mL) just dilutes the KHP solution but does not change the number of moles of KHP. So, the volume of NaOH required for the titration and thus its calculated molarity remain the same.
2. For the second part, in an acid - base titration using phenolphthalein as an indicator, the color change to a faint pink indicates the endpoint. If the color is dark pink, it means an excess of NaOH has been added. The endpoint is the point where the moles of acid (KHP) and base (NaOH) are in stoichiometric proportion. A dark pink color implies that the titration has gone past the equivalence point, leading to an inaccurate determination of the molarity of the base.

Answer:

1. The molarity of NaOH would not be affected because the moles of KHP, which determine the reaction with NaOH, do not change with the volume of water used to dissolve KHP.
2. The color change should be a faint pink because a dark pink color indicates an excess of NaOH and that the titration has passed the equivalence point.