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nomore asked in 科學及數學化學 · 1 decade ago

acid and alkaline

If strong acid is mixed with weak alkaline, why the pH value after neutralization is smaller than 7 but not 7? Then if excess weak alkaline is added to a strong acid, is that the pH is still smaller than 7? Why? Thanks so much!

Update:

then why when a strong acid is titrated against a weak alkaline, phenophthalein indicator will change to red? and why volume of acid required to neutralize a fixed volume of strong alkaline is the same regardless of whether it is a strong or weak acid?

Update 2:

I understand that if excess OH- is present , then the solution is alkaline. However, from the book, it mentioned that the pH is smaller than 7 after neutralization for strong acid VS weak alkaline (because the salt itself is acidic?)

Update 3:

Then if excess weak OH- is added, why pH of the solution will change from <7 to >7? is that only for the case of very excess OH-?

1 Answer

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  • 1 decade ago
    Favorite Answer

    1. Suppose 25ml 0.1M HCl vs 25ml 0.1M NH3 , indicator is methyl orange, colour change at pH 4.4 (yellow) to pH 3(red)

    Since strong acid is completely ionized, so 0.05ml (1drop) of 0.1M HCl is exceed, it can provide 0.05*0.1/1000 = 0.000005 mole H+ , [H+] = 0.0000999M, pH=-log[H+] = 4. Therefore the indicator colour change to orange. End point is detected.

    Excess alkaline is added, the pH will greater than 7. Since when all H+ is used up, then excess OH- is remained.

    The colour change range of Phenolphthalein is colourless (pH 8.2) to pink (pH 10). If the pH of weak alkali can reach pH 10, then pink colour can be observed.

    but if Na2CO3 is used, problem is happened. Since pH of Na2CO3 is > 10, but NaHCO3 is < 10; therefore Phenolphthalein cannot be used to titrate Na2CO3 against HCl.

    Strong mean the acid/alkaline is completely ionized in water.

    Weak mean the acid/alkaline is partly ionized in water.

    eg. Ethanoic acid in water, some few is ionized into CH3COO- and H+ form, most are still present as CH3COOH form.

    Source(s): me
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