a) Exactly 50.00 mL of a weak acid HA is titrated with 0.1000M NaOH. If the equi
ID: 531723 • Letter: A
Question
a) Exactly 50.00 mL of a weak acid HA is titrated with 0.1000M NaOH. If the equivalence point is reached after 39.30 mL of base is added, what is the original concentration of the HA? b) If at 19.65 mL of base added the pH = 4.85, what is the Ka of the acid? a) Circle the weakest acid. H_2SeO_3 H_2TeO_3 H_2TeO_4 H_2SeO_4 b) Circle the strongest base Br^- ClCH_2NH_2 NH_3 CH_3NH_2 c) Circle the acid with the lowest pKa HIO_3 HClO_3 HClO_2 HIO The pK_a of formic acid, CHOOH, is 3.72 and its K_a = 1.9 times 10^-4 (a) How many grams of solid sodium formate (CHOONa) must be added to 0.0800 M formic acid to make a buffer of pH = 3.50?Explanation / Answer
answer 7.. C1V1=C2V2
C1X50= 0.1000X39.30
C1=0.01000X39.30/50 ===0.0786 M
8..a) option B is correct, electronegaativity of central atom increase acidity increase while no .of oxygen atom increase H donating power increase. according to statement order follow like = h2seo4>h2seo3>h2teo4>h2teo3
b) option D is correct . CH3 is +I donating power which increase donating power of NH3 of loan pair.
c) option A is correct ,because HIO3 is strongest acid which have lowest pka value. as pka value increase acidity decrease while basicity increase.
9.. pH= pKa+log [salt/acid] given pH= 3.50 pKa= 3.72 acid concentration 0.0800 M find =salt concentration so now
3.50 =3.72 +log[salt/0.0800]
{-0.22} =log [salt/0.0800]
[salt/0.0800] =antilog10-0.22
[salt/0.0800]=0.6025 =0.0800x0.6025
salt=0.0482 M
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