Question

What is the molarity of sodium hydroxide solution?

A 0.3178 g sample of potassium hydrogen phthalate is dissolved in 100 mL of water. 33.75 mL of sodium hydroxide solution is required to reach the equivalence point.

Answer 1

I get `4.62*10^-2 \ "M"`.

Explanation:

I'll assume that one mole of sodium hydroxide is required to neutralize one mole of potassium hydrogen phthalate, since I don't think hydrogen phthalate exists by itself!

We first find the concentration (molarity) of the potassium hydrogen phthalate.

Potassium hydrogen phthalate has a molar mass of `204.22 \ "g/mol"`. So here, there exist:

`(0.3178color(red)cancelcolor(black)"g")/(204.22color(red)cancelcolor(black)"g""/mol")~~1.56*10^-3 \ "mol"`

And so, the molarity of this solution is:

`(1.56*10^-3 \ "mol")/(0.1 \ "L")=1.56*10^-2 \ "M"`

Now, we use the dilution equation, which states that,

`V_1c_1=V_2c_2`

• `V_1,V_2` are the volumes of the two solutions

• `c_1,c_2` are the concentrations of the two solutions

So we got:

`1.56*10^-2 \ "M"*100 \ "mL"=33.75 \ "mL"*c_2`

`c_2=(1.56*10^-2 \ "M"*100color(red)cancelcolor(black)"mL")/(33.75color(red)cancelcolor(black)"mL")`

`=4.62*10^-2 \ "M"`

EXTRA:

The `"pOH"` of a solution is given by: `"pOH"=-log[OH^-]`

• `[OH^-]` is the hydroxide ion concentration in terms of molarity.

So here,

`-log[4.62*10^-2]="pOH"`

`"pOH"~~1.34`

And so,

`"pH"=14-1.34=12.66`