반응형 일반화학/[16장] 산-염기 평형481 0.200 M HCl 10.00 mL + 0.250 M NaOH. 2.00 mL 0.200 M HCl 10.00 mL + 0.250 M NaOH. 2.00 mL In a titration, 10.00 mL of 0.200 M HCl(aq) is titrated with standardized 0.250 M NaOH(aq). What is the amount of unreacted HCl(aq) and the pH of the solution after the following volumes of NaOH(aq) have been added? b) 2.00 mL --------------------------------------------------- ▶ 참고: 강산-강염기 적정 [ https://ywpop.tistory.com/2732 ] -----------------------.. 2023. 1. 2. 0.200 M HCl 10.00 mL + 0.250 M NaOH. 0 mL 0.200 M HCl 10.00 mL + 0.250 M NaOH. 0 mL In a titration, 10.00 mL of 0.200 M HCl(aq) is titrated with standardized 0.250 M NaOH(aq). What is the amount of unreacted HCl(aq) and the pH of the solution after the following volumes of NaOH(aq) have been added? a) 0 mL --------------------------------------------------- ▶ 참고: 강산-강염기 적정 [ https://ywpop.tistory.com/2732 ] -----------------------------.. 2023. 1. 2. 짝산 짝염기. H3O^+ + S^2- ⇌ HS^- + H2O 짝산 짝염기. H3O^+ + S^2- ⇌ HS^- + H2O Complete the following reaction, identify acids and bases, along with the conjugate acid-base pairs. H3O^+ + S^2- ⇌ --------------------------------------------------- ▶ 참고: 짝산 짝염기 쌍 [ https://ywpop.tistory.com/2700 ] --------------------------------------------------- H3O^+ + S^2- ⇌ HS^- + H2O H3O^+(aq) + S^2-(aq) ⇌ HS^-(aq) + H2O(l) > 산: H3O^+ > 짝염기: H2O > 염기:.. 2022. 12. 30. 0.1 N 황산 용액의 pH 0.1 N 황산 용액의 pH 0.05 M 황산 용액의 pH --------------------------------------------------- ▶ 참고: 0.1 M 황산 용액의 pH [ https://ywpop.tistory.com/5813 ] --------------------------------------------------- H2SO4의 당량수 = 2 eq/mol 이므로, 0.1 N 황산 = 0.05 M 황산 ( 참고 https://ywpop.tistory.com/3093 ) 문제에서 황산의 Ka 값이 주어지지 않았기에... 황산의 2개 수소 전부 100% 이온화된다고 가정하면, [H^+] = 2 × 0.05 M = 0.1 M pH = –log[H^+] = –log(0.1) = 1 답.. 2022. 12. 26. 반-당량점. 0.054 M HNO2 KOH [H^+] 반-당량점. 0.054 M HNO2 KOH [H^+] A 0.054 M HNO2 solution is titrated with a KOH solution. What is the [H^+] at half way to the equivalence point? --------------------------------------------------- > HNO2의 Ka = 4.5×10^(-4) ( 참고 https://ywpop.tistory.com/22831 ) 반-당량점에서 용액의 pH = pKa 이므로, ( 참고 https://ywpop.tistory.com/10228 ) 반-당량점에서 [H^+] = Ka = 4.5×10^(-4) 답: 4.5×10^(-4) M 2022. 12. 19. 당량점 pH. 1 M HNO2 50 mL + 0.850 M NaOH 당량점 pH. 1 M HNO2 50 mL + 0.850 M NaOH Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution b) at half-neutralization? c) at the equivalence point? --------------------------------------------------- ▶ 참고: 약산-강염기 적정 [ https://ywpop.tistory.com/2736 ] --------------------------------------------------- 이런 문제에서는 약산의 Ka 값이 주어져야 한다. [참고] 대표적인 HNO2의 Ka = 4.5×.. 2022. 12. 17. 흰인 10.0 g 연소 500 mL 용액 25℃ 용액의 pH 흰인 10.0 g 연소 500 mL 용액 25℃ 용액의 pH 10.0 g white phosphorus burn water 500 mL solution pH 흰인 10.0 g이 충분한 산소 하에서 연소되었다. 생성물을 충분한 물에 녹여서 500 mL 용액을 만들었다면 25℃에서 이 용액의 pH는 얼마이겠는가? A 10.0 g sample of white phosphorus was burned in an excess of oxygen. The product was dissolved in enough water to make 500 mL of solution. Calculate the pH of the solution at 25°C. A 10.0 g sample of white phosphorus, P4 w.. 2022. 12. 15. 5.0 g white phosphorus burn excess oxygen water 250.0 mL solution 5.0 g white phosphorus burn excess oxygen water 250.0 mL solution A 5.0 g quantity of white phosphorus was burned in an excess of oxygen and the product was dissolved in water to make 250.0 mL solution. a) Write balanced equations for the reaction. --------------------------------------------------- P4 연소 반응식. 흰인 연소 반응식 P4 + 5O2 → P4O10 H3PO4 생성 반응식 P4O10 + 6H2O → 4H3PO4 [참고] H2CO3 생성 반응식 CO2 + .. 2022. 12. 15. 2.0×10^(-6) M H^+ 용액의 pH 2.0×10^(-6) M H^+ 용액의 pH 25℃에서 H^+의 농도가 2.0×10^(-6) M인 수용액의 pH는? --------------------------------------------------- pH = –log[H^+] ( 참고 https://ywpop.tistory.com/2706 ) = –log(2.0×10^(-6)) = 5.7 답: pH = 5.7 [참고] log(2.0) = 0.30 을 사용해서 계산하면, –log(2.0×10^(-6)) = –log(2.0) + –log(10^(-6)) = –0.30 + 6 = 5.7 ( 참고 https://ywpop.tistory.com/16776 ) 2022. 12. 13. Why is NH3 not regarded as an Arrhenius base? Why is NH3 not regarded as an Arrhenius base? Why is NH3 not considered a base under the Arrhenius definition of acids and bases? a) It does not react with acids. b) NH3(aq) does not contain OH^- ions. c) NH3 is considered a base under the Arrhenius definition, but not under the Bronsted-Lowry definition. d) It does not cause a color change when mixed with an acid-base (pH) indicator. ----------.. 2022. 12. 10. 이전 1 ··· 4 5 6 7 8 9 10 ··· 49 다음 반응형