Question

you prepare the standard solution from 0.300 g of K H P K H P in 30.0 ml of water. if the naoh solution is approximately 0.1 m, what is the approximate volume (in ml) of naoh solution you will need for the titration?

Answer 1

To determine the volume of NaOH needed for titration, the moles of KHP are calculated using its molar mass, and then used to find the volume of NaOH solution based on its concentration. For 0.300 g KHP, approximately 14.7 mL of 0.1 M NaOH is required.

Step-by-step explanation:

The student's question pertains to determining the approximate volume of NaOH solution needed for the titration of a standard solution prepared from 0.300 g of potassium hydrogen phthalate (KHP). To find this volume, one would need the molar mass of KHP and the stoichiometry of the reaction between KHP and NaOH. The balanced chemical equation for the reaction is:

KHC8H4O4(aq) + NaOH(aq) → NaKC8H4O4(aq) + H2O(l)

Molar mass of KHP (KHC8H4O4) = 204.22 g/mol

Therefore, moles of KHP = 0.300 g / 204.22 g/mol = 0.00147 mol

From the balanced equation, the mole ratio of KHP to NaOH is 1:1. Hence, 0.00147 mol of KHP will react with 0.00147 mol of NaOH. Given that the NaOH solution is approximately 0.1 M (molar), the volume (V) needed can be calculated by:

V = moles of NaOH / concentration of NaOH = 0.00147 mol / 0.1 mol/L = 0.0147 L = 14.7 mL

Thus, approximately 14.7 mL of a 0.1 M NaOH solution would be required for the titration.