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

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.