반응형 일반화학/[17장] 수용액 평형의 다른 관점400 0.1 M KCN 용액의 가수분해도, pH 0.1 M KCN 용액의 가수분해도, pH 단, HCN의 Ka = 1.7×10^(-5) --------------------------------------------------- 강전해질인 가용성염의 이온화 KCN(aq) → K^+(aq) + CN^-(aq) ---> [KCN] = [CN^-] = 0.1 M 약산(HCN)의 짝염기인 CN^-의 가수분해 CN^-(aq) + H2O(l) ⇌ HCN(aq) + OH^-(aq) ( 참고: 가수분해 https://ywpop.tistory.com/5502 ) 가수분해상수 Kh = [HCN] [OH^-] / [CN^-] = Kw / Ka = (10^(-14)) / (1.7×10^(-5)) = 5.88×10^(-10) Kh = (x) (x) / (C–x) ≒ x^2 .. 2023. 6. 14. 0.01 M NH4NO3 용액의 가수분해도, 가수분해상수, pH 0.01 M NH4NO3 용액의 가수분해도, 가수분해상수, pH 단, NH3의 Kb = 1.75×10^(-5) --------------------------------------------------- 강전해질인 가용성염의 이온화 NH4NO3(aq) → NH4^+(aq) + NO3^-(aq) ---> [NH4NO3] = [NH4^+] = 0.01 M 약염기(NH3)의 짝산인 NH4^+의 가수분해 NH4^+(aq) + H2O(l) ⇌ NH3(aq) + H3O^+(aq) ( 참고: 가수분해 https://ywpop.tistory.com/5502 ) 가수분해상수 Kh = [NH3] [H3O^+] / [NH4^+] = Kw / Kb = (10^(-14)) / (1.75×10^(-5)) = 5.71×10^(-1.. 2023. 6. 14. 0.1 M HAc 50 mL + 0.1 M NaOH 50 mL. 당량점 pH 0.1 M HAc 50 mL + 0.1 M NaOH 50 mL. 당량점 pH 약산-강염기 적정. 당량점 pH 0.1 M HAc 50 mL 중화에 0.1 M NaOH 50 mL 소비. 이 용액의 [H^+]? 단, HAc의 Ka = 1.8×10^(-5) --------------------------------------------------- ▶ 참고: 약산-강염기 적정 [ https://ywpop.tistory.com/2736 ] --------------------------------------------------- (0.1 mol/L) (50/1000 L) = 0.005 mol = HAc의 몰수 = NaOH의 몰수 약산의 몰수와 강염기의 몰수가 같으므로, 약산은 강염기에 의해 완전히 중화된다. H.. 2023. 6. 11. 0.0319 M 벤질아민 50.0 mL + 0.0500 M HCl 적정. Va = 30.0 0.0319 M 벤질아민 50.0 mL + 0.0500 M HCl 적정. Va = 30.0 0.0319 M 벤질아민 용액 50.0 mL를 0.0500 M HCl로 적정 Va = 30.0 에서, 용액의 pH A 50.0 mL solution of 0.0319 M benzylamine was titrated with 0.0500 M HCl. Calculate the pH at the following volumes of added acid: Va = 0, 12.0, 1/2Ve, 30.0, Ve, 35.0 mL. pKb of benzylamine is 4.66; Ve = Volume at equivalence point. -------------------------------------------------.. 2023. 6. 8. 0.0319 M 벤질아민 50.0 mL + 0.0500 M HCl 적정. Va = 1/2Ve 0.0319 M 벤질아민 50.0 mL + 0.0500 M HCl 적정. Va = 1/2Ve 0.0319 M 벤질아민 용액 50.0 mL를 0.0500 M HCl로 적정 Va = 1/2Ve 에서, 용액의 pH A 50.0 mL solution of 0.0319 M benzylamine was titrated with 0.0500 M HCl. Calculate the pH at the following volumes of added acid: Va = 0, 12.0, 1/2Ve, 30.0, Ve, 35.0 mL. pKb of benzylamine is 4.66; Ve = Volume at equivalence point. -----------------------------------------------.. 2023. 6. 7. 0.0319 M 벤질아민 50.0 mL + 0.0500 M HCl 적정. Va = 12.0 0.0319 M 벤질아민 50.0 mL + 0.0500 M HCl 적정. Va = 12.0 0.0319 M 벤질아민 용액 50.0 mL를 0.0500 M HCl로 적정 Va = 12.0 에서, 용액의 pH A 50.0 mL solution of 0.0319 M benzylamine was titrated with 0.0500 M HCl. Calculate the pH at the following volumes of added acid: Va = 0, 12.0, 1/2Ve, 30.0, Ve, 35.0 mL. pKb of benzylamine is 4.66; Ve = Volume at equivalence point. -------------------------------------------------.. 2023. 6. 7. 0.0319 M 벤질아민 50.0 mL + 0.0500 M HCl 적정. Va = 0 0.0319 M 벤질아민 50.0 mL + 0.0500 M HCl 적정. Va = 0 0.0319 M 벤질아민 용액 50.0 mL를 0.0500 M HCl로 적정 Va = 0 에서, 용액의 pH A 50.0 mL solution of 0.0319 M benzylamine was titrated with 0.0500 M HCl. Calculate the pH at the following volumes of added acid: Va = 0, 12.0, 1/2Ve, 30.0, Ve, 35.0 mL. pKb of benzylamine is 4.66; Ve = Volume at equivalence point. Calculate the pH at the following volumes of added acid.. 2023. 6. 7. HC2H3O2(aq) pKa 4.76 100.0 mL 0.300 M NaC2H3O2(aq) 0.100 M NaOH(aq) HC2H3O2(aq) pKa 4.76 100.0 mL 0.300 M NaC2H3O2(aq) 0.100 M NaOH(aq) HC2H3O2(aq)의 pKa는 4.76이다. 100.0 mL의 0.300 M NaC2H3O2(aq), 100.0 mL의 0.300 M HC2H3O2(aq), 그리고 100.0 mL의 0.100 M NaOH(aq) 를 포함하고 있는 용액의 pH는 얼마인가? --------------------------------------------------- ▶ 참고: 아세트산 완충용액에 강산 또는 강염기를 가했을 때 [ https://ywpop.tistory.com/4319 ] --------------------------------------------------- > 약산의 몰수 = (0.. 2023. 6. 5. solubility of CaF2 at pH 3.0 is 3.86×10^(-x) M solubility of CaF2 at pH 3.0 is 3.86×10^(-x) M The solubility of CaF2 in a solution buffered at pH = 3.0 is 3.86×10^(-x) M. Calculate the value of x. Ka for HF = 6.3×10^(-4) and Ksp of CaF2 = 3.45×10^(-11) --------------------------------------------------- ▶ 참고: Ksp와 몰용해도 [ https://ywpop.tistory.com/8434 ] --------------------------------------------------- CaF2(s) ⇌ Ca^2+(aq) + 2F^-(aq) Ksp = .. 2023. 6. 3. 0.05 M pyruvic acid + 0.06 M sodium pyruvate buffer pH 0.05 M pyruvic acid + 0.06 M sodium pyruvate buffer pH Find the pH of a buffer solution with pyruvic acid and sodium pyruvate, given the concentration of pyruvic acid was 0.05 M and the concentration of sodium pyruvate was 0.06 M. Find the pH of a buffer made by mixing 0.05 M pyruvic acid with 0.06 M sodium pyruvate. Ka for pyruvic acid = 3.1×10^(-3). 단, pyruvic acid의 Ka = 3.1×10^(-3) ----------.. 2023. 6. 3. 이전 1 2 3 4 5 6 7 ··· 40 다음 반응형
