Question

Use the Henderson-Hasselbalch equation to calculate the pH of each solution: a solution that contains 0.695% C5H5N by mass and 0.845% C5H5NHCI by mass Express your answer using two decimal places. Hν ΑΣφ ? рH- 15.2 Previous Answers Request Answer Submit Incorrect; Try Again; One attempt remaining Part C a solution that is 13.5 g of HF and 24.0 g of NaF in 125 mL of solution Express your answer using two decimal places. ΑΣφ ? pH Request Answer Submit

Answer 1

Answena a bffer otem by mixing CsHs N CsH5NHcl pa o1. 7iven and given, mass % ot esHsN = d.695% mass oaCs HsNHel = 0-845% it 100 sotution i thee. 0-6957 of esHs N79 g/mol Cs HSN mass MolaH mass mumbu moles a.a087mel. 79.0 g/mol CSHS NHC= 0.845 and massag Molan mass ot csHgNHCl = 115.5 a/mol number O.845 11S.5 a/mol moles = = d-0073mol. to Hendenson- Hasselbalch equation, [sall] L base acconding pHE pкъ + log [вн*] Kb fot CsHsN (pyridi'ne) 1.7 x 10-9 l kog7x10-7). = 8.76. insiead of [вн+] mo. moe ration beéause cam se Jor base voluene o he solution nemam eomtont a salt (conjnyalaaad). and moles of BH moles PH E pКь +og pH 8.76 Lomolecof wi t5 NH molec of esHsN pH 9.76 tlo7 d.0073 O.0087 at, pH=8-76 - 0,07 = 8-69. ( Answer).

19-0mol given mass of HF = 13.5; molan mass f HF= aiven Molar mass. 02. HF = mass no.o moles g a.71 mol Na Fs 24-07; molan mass of NaF = 23+ 19 42/mol. Ven mass 24 moles NaF = 0.S7 mol. 42 The 4ofution = Now, vofume 125mL = Q.125 L. 20.0fmoles Volume (L) concentration (Molanity o HF= .71 mol 0-125L 5.68 M Concentration (Mokanitg) o O.S7mol O.125 L Nafoi Molanity gNaF is same as Moloriity of Fa 4. $6M pH=PKa +lo pha for HF= 3-1. [4-SEJ PH 3.1oa [pe] 3-1 t lo [s9-S1 3.01. (Answer) pH 3.1 - 0.09 =