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# Acid–base titrations

In a titration, a solution of known concentration (the titrant) is added to a solution of the substance being studied (the analyte). In an acid-base titration, the titrant is a strong base or a strong acid, and the analyte is an acid or a base, respectively. The point in a titration when the titrant and analyte are present in stoichiometric amounts is called the equivalence point. This point coincides closely to the endpoint of the titration, which can be identified using an indicator. Created by Jay.

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• That MCAT does not allow for a calculator... how would you do this then?
• I find that writing everything in scientific notation is really helpful! For example, at you can do it like this:
.00486 mol / .02 L = 4.86 x 10^-3 / 2 x 10^-2
I like to separate it out like this (4.8 / 2) x 10^-3 - (-2)
2.4 x 10-1 = .24 M
• ***
Please correct me if I'm wrong
I've read that phenolphthalein changes color when the pH is from above the range of 8.3 to 10.0
Conclusions:
1: Using it we have the change in color when we have more OH- than those used to neutralize the strong acid.
2: As the range is from 8.3 to 10.0, we would have an error margin of about 2 decimal points in the total concentration of H+
• Clarification to conclusion 1: The solution will NOT change color at the exact instant that [OH-] first exceeds [H+], because the pH at this point would only be very slightly above 7 (which is not in the pH range of phenol). A color change will happen when [OH-] ions is about (2*10^-6 M), because this is when pH is at 8.3. The color change marks the end point of the titration. Hope this helps!
• What is the difference between end point and equivalence point?
• Great question! The equivalence point is almost more of a mathematical concept, it is when the moles of H+ in solution equals the moles of OH- in solution, and all the acid (or base) in the original solution is neutralized.

But how do we know when the equivalence point has been reached? We know we have reached the equivalence point by adding an acid-base indicator such as phenolphthalein, we look for when the solution turns pink. But as Jay mentions at , phenolphthalein turns pink when there is base present. If there is base present, that means we are at least a teensy bit past the equivalence point, since there isn't any base at the equivalence point. When the solution turns pink and we stop the titration, that is called the end point. But the endpoint is usually just a little bit past the equivalence point, because that is how the indicators work.
• does M (mol) means also N (nomality) ?
• M is Molarity (Not mol)
N=>Normality
N=M*n factor (For redox reactions)
N=M*acidity (For bases)
N=M*basicity (For acids)
• For the short cut equation does the molar ratio between acid and base matter. Would you need to need to do the calculation differently if for every 2 mol of compound A you need 3 mol of compound B to make the product?
• the equation we use here is CaVa/CbVb=na/nb
where Ca is concentration of acid and Va is volume of acid
Cb is concentration of base and Vb is volume of base
na/nb is the mole ratio of the acid and base
• isn't this amount not exact as there was excess of NaOH causing the solution to be pink. isn't the volume of HCl less than 0.00486 mol
• Which is correct Ca(OCl)Cl or CaOCl(subscript)2? And what is the chemical name of the compound?
• the correct formula is CaOCl(subscript)2

it is an inorganic compound called calcium hypochlorite or u may commonly know it as bleaching powder, used as a disinfectant. as it is a mixture of lime and calcium chloride, it is often called as chlorine powder.

hope this helps :)
• When calculating the [HCl], wouldn't the final volume be the initial 20 mL (from HCl) plus the additional 48.6 mL (NaOH)? And thus, the concentration would ~0.071 M instead of 0.243 M?
• You are calculating the concentration of HCl at the beginning of the titration, that is, when the HCl was in the original 20 mL.
• For the short cut equation, wouldn't you need to convert the mL to L?
• The short cut equation is known as the dilution formula.

For any equation we use in science, the only requirement regarding the units is that they agree with each other. What that means if that the units can be anything, as long as they are the same unit. So for example we can use mL for the volume so long as the other volume is also in mL; it wouldn't work if one was in mL and the other was in L.

If we write the formula using the units as the terms:
(M1)(V1) = (M2)(V2)
(M)(mL) = (M)(mL), both sides of the equation equal each other since they are identical. By the same logic we could also have the molarity in a different unit like millimolar, mM, and the formula still works as long as both concentrations have the same unit.

So even if the product of molarity and mL isn't something we're interested in, for the dilution formula it's allowed as long as the concentration and volume units are the same for both sides of the equation.

Hope that helps.