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Consider the titration of 25.0 mL of 0.175 M cyanic acid, HCNO, with 0.250 M LiOH....

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Chemistry

Consider the titration of 25.0 mL of 0.175 M cyanicacid, HCNO, with 0.250 M LiOH. The Ka of HCNO is 3.5 * 10^-4.

What is the volume of LIOH required to reach theequivalence point?
b. calculate the pH after the following volumes of LIOH have beenadded:
a. 0mL
b. 5.0 mL
c. 8.75 mL
d. 17.5 mL

Answer & Explanation Solved by verified expert
3.7 Ratings (448 Votes)
a find the volume of LiOH used to reach equivalence point MHCNOVHCNO MLiOHVLiOH 0175 M 250 mL 025M VLiOH VLiOH 175 mL Answer 175 mL b 1when 00 mL of LiOH is added HCNO dissociates as HCNO H CNO 0175 0 0 0175x x x Ka HCNOHCNO Ka xxcx Assuming x can be ignored as compared to c So above expression becomes Ka xxc so x sqrt Kac x sqrt 351040175 7826103 since x is comparable c our assumption is not correct we need to solve this using Quadratic equation Ka xxcx 35104 x20175x 6125105 35104 x x2 x2 35104 x6125105 0 This is    See Answer
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