[Queen09]

How to find the molarity of NaOH when the given is
0.1803 grams of KHPg/mol
Molecular w. of KHP is 204.2
9.7 ml NaOH

[Unknown008]

I'm not sure I understand your question.

You reacted 0.1803 g of KHP with 9.7 mL of NaOH for neutralisation?

The mole ratio is 1:1 from the chemical equation you have to set up.

The number of moles of KHP used is 0.1803/204.2 mol

This number of moles is equal to that of NaOH.

So, in 9.7 mL of NaOH, you have so much moles of NaOH.
How many moles of NaOH are there in 1000 mL?

This is also the molarity.

[Queen09]

oh, it's okay now,
KHP + NaOH -> KNaP + H2O

if 0.1803 g of KHP has been used, this is

0.1803 g/204.2 g mol^-1
=8.829 x 10^-1 mole

which reacts with the same number of moles of NaOH

so concentration of NaOH is

8.829 x 10^-1 mole /0.0097 L
=0.091 M

thank you :)

maybe you can help me here:
an antacid preparation of Na2CO3 claims that a 3.00 gram tablet contains enough antacid to neutralize full stomach acid (0.100 M HCl). Assuming that the average stomach acid contains 0.750 L of acid, calsulate the moles of Na2CO3 in the antacid.

[Unknown008]

1. Find the number of moles of HCl which you need to neutralise.
2. Write down the equation for the reaction between HCl and Na2CO3.
3. Balance and deduce the mole ratio.
4. This amount of moles is present in the tablet of antacid.

The question doesn't seem complete though... maybe it then asks you to show that the brand is lying or not?