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일반화학/[16장] 산-염기 평형402

pH 4.5인 물의 수소이온농도(M) pH 4.5인 물의 수소이온농도(M) pH = 4.5 인 물의 수소 이온 농도(M)는? --------------------------------------------------- pH = –log[H^+] = 4.5 이므로, ( 참고 https://ywpop.tistory.com/2706 ) [H^+] = 10^(-4.5) M = 3.16×10^(-5) M 답: 약 3.2×10^(-5) M 2023. 2. 1.
수용액 안의 수소 이온 농도 수용액 안의 수소 이온 농도 1. 아세트산이 녹은 물의 수소이온 농도는 순수보다 높은가 낮은가? 2. 에틸아민이 녹은 물의 수소이온 농도는 순수보다 높은가 낮은가? --------------------------------------------------- 순수 안의 수소 이온 농도, [H^+] = 1×10^(-7) M ( 참고 https://ywpop.tistory.com/2706 ) 1. 아세트산이 녹은 물의 수소이온 농도는 순수보다 높은가 낮은가? ---> 아세트산은 산이므로, 순수보다 높다. ( 참고 https://ywpop.tistory.com/18330 ) 2. 에틸아민이 녹은 물의 수소이온 농도는 순수보다 높은가 낮은가? ---> 에틸아민은 염기이므로, 순수보다 낮다. ( 참고 https:/.. 2023. 1. 20.
pH 4.2 buffer 1.0 M HC2H3O2 1.0 L NaC2H3O2 NaOH pH 4.2 buffer 1.0 M HC2H3O2 1.0 L NaC2H3O2 NaOH pH 4.2 완충 용액 제조에 필요한 물질 a. 1.0 M HC2H3O2 1.0 L에 몇 몰 NaC2H3O2? b. 1.0 M HC2H3O2 1.0 L에 몇 몰 NaOH? 단, HC2H3O2의 Ka = 1.8×10^(-5) --------------------------------------------------- Henderson-Hasselbalch 식 pH = pKa + log([짝염기]/[약산]) ( 참고 https://ywpop.tistory.com/1926 ) 약산, HC2H3O2의 몰수 = (1.0 mol/L) (1.0 L) = 1.0 mol pKa = –logKa = –log(1.8×10^(-5)) = 4.. 2023. 1. 16.
Kb 3.5×10^(-6) 0.24 M weak base pH Kb 3.5×10^(-6) 0.24 M weak base pH Calculate the pH of a 0.24 M solution of a weak base with a Kb of 3.5×10^(-6). --------------------------------------------------- Kb = x^2 / (C–x) ( 참고 https://ywpop.tistory.com/4294 ) (C–x) ≒ C 라 근사처리하면, x = [Kb × C]^(1/2) = [(3.5×10^(-6)) × 0.24]^(1/2) = 0.0009165 M = [OH^-] pH = 14 – pOH = 14 – (–log(0.0009165)) = 10.96 ( 참고 https://ywpop.tistory.com/2706 ).. 2023. 1. 13.
0.105 M CH3COOH 24.0 mL + 0.130 M NaOH. 당량점 pH 0.105 M CH3COOH 24.0 mL + 0.130 M NaOH. 당량점 pH 24.0 mL of 0.105 M acetic acid (HC2H3O2) will be titrated with 0.130 M sodium hydroxide (NaOH). The Ka of acetic acid is 1.77×10^(-5). e) The pH at the equivalence point --------------------------------------------------- ▶ 참고: 약산-강염기 적정 [ https://ywpop.tistory.com/2736 ] --------------------------------------------------- CH3COOH의 초기 몰수 = (0.105 mo.. 2023. 1. 6.
0.105 M CH3COOH 24.0 mL + 0.130 M NaOH. 반-당량점 pH 0.105 M CH3COOH 24.0 mL + 0.130 M NaOH. 반-당량점 pH 24.0 mL of 0.105 M acetic acid (HC2H3O2) will be titrated with 0.130 M sodium hydroxide (NaOH). The Ka of acetic acid is 1.77×10^(-5). d) The pH at half-equivalence point (= after adding one-half of the NaOH volume calculated in question b) --------------------------------------------------- ▶ 참고: 약산-강염기 적정 [ https://ywpop.tistory.com/2736 ] -----.. 2023. 1. 4.
0.105 M CH3COOH 24.0 mL + 0.130 M NaOH. 6.00 mL pH 0.105 M CH3COOH 24.0 mL + 0.130 M NaOH. 6.00 mL pH 24.0 mL of 0.105 M acetic acid (HC2H3O2) will be titrated with 0.130 M sodium hydroxide (NaOH). The Ka of acetic acid is 1.77×10^(-5). c) The pH after adding 6.00 mL of NaOH Consider the titration of a 24.0 mL sample of 0.105 M CH3COOH with 0.130 M NaOH. What is the pH at 6.00 mL of added base? ---------------------------------------------------.. 2023. 1. 4.
0.105 M CH3COOH 24.0 mL + 0.130 M NaOH. 당량점 부피 0.105 M CH3COOH 24.0 mL + 0.130 M NaOH. 당량점 부피 24.0 mL of 0.105 M acetic acid (HC2H3O2) will be titrated with 0.130 M sodium hydroxide (NaOH). The Ka of acetic acid is 1.77×10^(-5). b) the volume of added base required to reach the equivalence point Consider the titration of a 24.0 mL sample of 0.105 M CH3COOH with 0.130 M NaOH. What is the volume of added base required to reach the equivalence .. 2023. 1. 4.
0.105 M CH3COOH 24.0 mL + 0.130 M NaOH. initial pH 0.105 M CH3COOH 24.0 mL + 0.130 M NaOH. initial pH 24.0 mL of 0.105 M acetic acid (HC2H3O2) will be titrated with 0.130 M sodium hydroxide (NaOH). The Ka of acetic acid is 1.77×10^(-5). a) the initial pH Consider the titration of a 24.0 mL sample of 0.105 M CH3COOH with 0.130 M NaOH. What is the initial pH? --------------------------------------------------- ▶ 참고: 약산-강염기 적정 [ https://ywpop.tisto.. 2023. 1. 4.
0.200 M HCl 10.00 mL + 0.250 M NaOH. 8.00 mL 0.200 M HCl 10.00 mL + 0.250 M NaOH. 8.00 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? c) 8.00 mL --------------------------------------------------- ▶ 참고: 강산-강염기 적정 [ https://ywpop.tistory.com/2732 ] -----------------------.. 2023. 1. 2.
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