Calculate the theoretical amount of .5M NaOH that it would take to neutralize approx. 2.0 g KHP. KHP is a monoprotic acid, the molar mass of KHP is 204.22 g/mol and the molar mass of NaOH is 40.00 g/mol. Answer in terms of mL.

I converted g KHP to mols, then mols of KHP to mols of NaOH. Then I converted mols of NaOH to g of NaOH and used the molarity equation (M=mols/liters) to get the volume of of NaOH. I got .783L or 783 mL is this correct?

The eyes of Texas are upon you, Brad

No. Your error is you convert mols NaOH to grams NaOH then used GRAMS for mols in M = mols/L but you can't do that. You should convert 2 g to mols KHP, then to mols NaOH, then mols NaOH/M = L NaOH

NaOH + KHPh => NaKPh + HOH

Need (2/204.22)mole NaOH = 0.0098-mole NaOH b/c Rxn is 1:1 Rxn Ratio. Ask what volume of 0.50M NaOH would deliver (2/204.22)mole of NaOH to neutralize 2g KHPh or 0.0098 mole KHPh? That is, (0.50M NaOH)(Volume of NaOH soln)= 0.0098 mole of NaOH delivered to neutralize 2 gms KHPh. Solve for Volume = 0.0196-Liter = 19.6 ml of the 0.5M NaOH(aq) solution will neutralize 2-gms KHPh (=0.0098 mole KHPh neutralized).

Well, well, well, looks like we have a chemistry whiz here! Let's test those calculations, shall we?

To start, we need to find the number of moles of KHP. Given that the molar mass of KHP is 204.22 g/mol and we have 2.0 g, we simply divide 2.0 g by 204.22 g/mol. This gives us approximately 0.0098 moles of KHP.

Since KHP is monoprotic, it means that it requires an equal number of moles of NaOH for neutralization. So, we have 0.0098 moles of NaOH.

Now, to find the amount of NaOH in grams, we multiply the number of moles (0.0098 moles) by the molar mass of NaOH (40.00 g/mol). This gives us 0.392 g of NaOH.

So far, so good?

Now, let's move on to finding the volume of NaOH using the molarity equation, as you mentioned. We have the moles of NaOH (0.0098 moles), and we know that the molarity is 0.5 M. By rearranging the equation M = moles/volume, we can isolate the volume and solve for it. Plugging in the values, we get:

0.5 M = 0.0098 moles / volume

Rearranging the equation, we find that the volume is approximately 0.0196 L, or 19.6 mL.

Uh-oh, it seems like our calculations took a different turn. According to my calculations (and mathematically-backed comedy), it seems that you may have gone wrong somewhere.

The correct answer is approximately 19.6 mL of 0.5 M NaOH required to neutralize approximately 2.0 g of KHP. So, it seems we've got ourselves a circus of calculations. But don't worry, calculating can be a bit of a balancing act sometimes!

To calculate the theoretical amount of 0.5 M NaOH needed to neutralize approximately 2.0 g of KHP (potassium hydrogen phthalate), you correctly followed the steps of converting grams to moles, then moles to grams, and finally using the molarity equation to find the volume in liters. Let's go through the calculation to double-check your answer:

1. Convert grams of KHP to moles:
2.0 g KHP x (1 mol KHP / 204.22 g KHP) = 0.00978 mol KHP

2. Since KHP is monoprotic (dissociates in water to release one H+ ion), the moles of KHP will be equal to the moles of NaOH needed for neutralization.

3. Convert moles of NaOH to grams:
0.00978 mol NaOH x (40.00 g NaOH / 1 mol NaOH) = 0.391 g NaOH

4. Use the molarity formula to find the volume of NaOH:
Molarity = moles / volume (in liters)

0.5 M NaOH = 0.00978 mol NaOH / Volume

Rearranging the formula to solve for the volume:
Volume = moles / Molarity = 0.00978 mol NaOH / 0.5 M NaOH = 0.01956 L

5. Convert liters to milliliters:
0.01956 L x (1000 mL / 1 L) = 19.56 mL

Therefore, the theoretical amount of 0.5 M NaOH required to neutralize approximately 2.0 g of KHP is 19.56 mL, not 783 mL. Your calculation was incorrect, but if you follow the steps described here, you should arrive at the correct answer.