Concentration of the sodium hydroxide solution?

Question

A solution of KHP (molar mass= 204.2) is made by dissolving 2.5525g og KHP in 250.0mL of water.The titration of 25.00mL of the KHP solution requires 40.00mL of a sodium hydroxide solutionto reach phenolphtalein end point. What is the conc. of the sodium hydroxide solution.?

Pls,not only answer,clear explanation...thanks

Answer 1

Okay, remember Ci*Vi = Cf*Vf? Well, this sort of like that, except there's not exactly an initial and final concentration, but it will still work ^_^. Are you thrill to hear that? So, pretend it's like C1*V1 = C2*V2.

So...let's list all of our given info:
2.5525g of KHP dissolved in 250.0 mL of water
C1 = ???
V1 = 25.00 mL of KHP
C2 = ???
V2 = 40.00 mL of NaOH (sodium hydroxide)

Let's find the concentration of C1 or the concentration of KHP. What do we know about concentration? It's usually the entity of the "thing" divided by/over the volume of the "thing", right? In chemistry, they like concentration in molarity (M or mole/L)

Let's convert the grams of KHP into moles of KHP. To do so, we'll divide the given grams of KHP by it's molar mass.

mole of KHP = 2.5525g KHP * (1 mole of KHP / 204.2g KHP) = 0.01250 mole of KHP [note: 4 sig.fig]

Now, let's convert the 250.0mL to L [that should be easy]
250.0mL * (1L / 1000mL) = 0.2500 L [note: 4 sig.fig]

C1 = 0.01250 mole of KHP / 0.2500 L = 0.05000 mole of KHP/L or 0.05000 M


Rearrange the given formula to solve fo C2. Yep, just plug them in the right place!
C2 = (C1*V1) / V2

C2 = (0.05000M * 25.00mL) / 40.00mL
C2 = 0.03125 M

Answer 2

equivalents of KHP= weight/eq. weight
=2.5525/204.2
=0.0125
normality= eq of KHP/volume
=0.0125/0.250
=0.05 N
On titration equivalents of KHP= Eq of NaOH

0.05*25 = x*40
x = 0.03125
Conc of NaOH=0.03125 N

Answer 3

calculate moles KHP used
2.5525g/(204.2g/mol)= 0.0125 mol KHP

calculate concentration KHP soln
[KHP]= 0.0125 mol /(0.25L) = 0.05M KHP

calculate moles in 25ml of KHP solution
?mol KHP = (0.025L KHP soln)(0.05mol/L) = 0.00125mol KHP used

Asssuming a 1:1 KHP:NaOH neutralization ratio,

this means there were 0.00125mol NaOH used to
neutralize the KHP

[NaOH] = (0.00125mol NaOH)/(0.04L) =
0.3125M NaOH