This is a question of the chemistry book with solution, however, there were seve

Question

This is a question of the chemistry book with solution, however, there were several points that I don’t understand(which I emphasized with red pen) Chapter 18-Acid-Base Equilbria Calculating Equilibrium Concent Acid for a Polyprotic SAMPLE PROBLEM 18.10 Using the egailibeium HA" 5x10°. ) found in citrus fruit IHAse 1. LAse and the pH of 0.050 H,Asc. Plan We kmow the initial con 0050 M) and both K,'s for -itaalconce, rancentrations of all species and convene, >> Ka2, so it make only in the second o check procestrations excepet (M,Asel FOLLOW-UP PROBLEMS 18.10A Oxalic acid (HoOC- Check Ki PH. We finst write the LO-I-HA".. Aho, because A., issmall, the amount ofHdet0L carpe neglected. We set up a reaction table for the first (H,Asle and then we solve for IH,O" and HAsc 1. Because Sc *sume that the first dissociation produces almost all the Horau a- H Writing the equations andesperessions check whether you need the 18.108 Carbonie acid (H.c the pH of a 0.150ibri [HAS THO*)-1.0×10-5 acidification. Using i [11,Asc] SOME SIMILAR PROBLEMs 5X10 Seting up a reaction table with xHAscaHAs H,o' Concentration (M) HAsclag)+ H,OUH,olag)+ Summary of Sect Two common types We simplify the arith H,O m HA and can 0.050 ta 0.050-x The fraction of wea solution, even thou Polyprotic acids ha dissociation provid Making the assumpticns: 2. Because K, is small, [H,Asc[HAse H,Asclir Thus H Asc] -0.00 M-0.050M MOL AND 18.5 Substituting into the expression for K, and solving for x The strength of an turn on the strengt in atomic and bo hydrides and oxo ILAsc] 0050 x [HAsc . ]> [HO*) ~ 7.1×10-4 M pH=-btHO.] =-log (7.110")-3.15 Checking the assumptions: H,0 T tFor any second dissociatin that do. Acid Strengt Two factors deta X10 · The electron The strength This is even less than H,Oer so the assumption is justified. Figure 18.9 dis 7.1x10AM 1. Across 0.050 M-X100-1.4%(5%; assumption E determines tive, it withdr a result, H' is solution, the increase in a is justified). Also, note that HAse0.050

Explanation / Answer

Here, Ka1 is the acid dissociation constant for H2ASc and Ka2 is the acid dissociation constant for HASc-. Therefore, equation for Ka2 should be,

Ka2 = [ASc2-][H3O+]/[HASc-]. So, it’s a typing error.

When you have comparable acid dissociation values, you need to calculate [H3O+] from both ionization reaction. But, here since Ka2 is too small, the corresponding dissociation will be too small and need not consider change of [H3O+] from the second ionization reaction.

Hope this helps. Use comment box if further clarifications required

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