posted by T on .
How many grams of KHP (204.22 g/mol) were present in container if 49.36 mL of 0.2264M NaOH were needed to reach the end point of a titration? I'm confused on the formula you would use for this.
Most problems in chemistry hinge on mols. A mole reacts with one/two/three/etc mols of some other material.
KHP + NaOH ==> NaKP + H2O
mols NaOH = M x L = ?
molsl KHP = mols NaOH (from the coefficients in the balanced equation).
Then mols KHP = grams KHP/molar mass KHP. You know molar mass and mols solve for grams KHP.
So, Mols NaOH = 0.2264 M NaOH x 0.04936 L NaOH = 0.011175
Mols KHP = grams KHP / molar mass KHP or
grams KHP = Mols KHP x molar mass KHP so
0.011175 x 204.22 = 2.2822 grams of KHP
Is that correct DBob222?
Yes, that is correct(almost); however, you have used too many significant figures. The 49.36 number limits the number of s.f. to four; therefore, you may have only four in the final answer. That means you should round your final answer to 2.282 g KHP.