A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M
HClO4 s...

Question

A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M 
HClO4 solution. Calculate the pH after the following volumes 
of acid have been added: (c) 24.0 mL.

Accepted Answer

Hello everyone today. We are being given the falling problem .175 moller of nitrite. It's used to titrate 25 millimeter sample of 250.35 molar sodium hydroxide. What is the P. H. F. For 12.5 mL of acid is added. So first you want to know a couple of things. H&O. three or nitrate is a strong acid and our sodium hydroxide is a strong base. Next we want to go ahead and find how many moles are based on how many moles of our acid that we have so far moles of acid. We're simply going to take our zero point 35175 moles per liter. That equals more clarity. We're gonna multiply that by our 25 ml. But we also must convert this into leaders without our units match up. And so we can see that one leader is equal to 1000 militia leaders. Sorry, units will cancel out here will be left with moles, will be left with 4.375 times 10 to the negative three moles of our acid From the base. We are going to do a very similar procedure. We're going to take our 0.350 mph, which also equals the number of moles. We're gonna multiply that by our 12.5 middle leaders since we are tight trading our base with our acid. And so we're going to once again convert this into leaders by saying that one leader is equal to 1000 middle leaders. When our units cancel out who will be left with 4.375 times 10 to the negative three. And we notice here that our moles of our acid are equal to our moles of our base. And so when that happens, this is when we have a an equivalence point met. So we have an equivalent point when the moles of asset equal the most of face. And so the equivalence point in this case is when Ph. is equal to seven. And so the Ph will be seven. I hope this helped. And until next time.

A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added: (c) 24.0 mL.

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