Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of naoh solution to reach the end point detected by phenolphthalein indicator. find the concentration of naoh expressed as mol naoh/kg solution.

1.062 mol/kg.

Step 1. Write the balanced equation for the neutralization.

MM = 204.22 40.00

KHC8H4O4 + NaOH → KNaC8H4O4 + H2O

Step 2. Calculate the moles of potassium hydrogen phthalate (KHP)

Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)

= 4.035 mmol KHP

Step 3. Calculate the moles of NaOH

Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)

= 4.035 mmol NaOH

Step 4. Calculate the mass of the NaOH

Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)

= 161 mg NaOH

Step 5. Calculate the mass of the water

Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g

= 37.973 g

Step 6. Calculate the molal concentration of the NaOH

b = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.