测定醋（稀醋酸）的Ka值 PASCO scientific1Chemistry Experiment Library: 06020 Science Work shop C31 Vinegar Experiment C31: Neutralization of Vinegar with Drain Cleaner (pH Sensor) (pH 传感器)Concept: acids, bases & salts概念：酸、碱和盐 Time: 30 m SW Interface: 300, 500 & 700 Macintosh_ file...
What is the pH when 30.0 mL of 0.10 M HCOOH solution is titrated with 8.6 mL of 0.20 M NaOH? (Ka = 1.8 x 10^-4) Given that Kb for (CH3)2NH is 5.4\times0-4 at 25C, what is the value of Ka for (CH3)2NH2 at 25 C? What is the Kb for the followin...
so now that i have my Ka value and my hydronium value i was considering making an "ICE" box to find my HA value but this is my conundrum: HCl + NaOH -> H2O + NaClwith this kind of equation i am working with my Ka should be NaCl/[HCl][NaOH]. but my Ka should look like this...
To calculate the pH of a 0.1 M acetic acid (CH₃COOH) solution with a given dissociation constant (Kₐ = 1.8 × 10⁻⁵), we can follow these steps:Step 1: Write the dissociation equation Acetic acid dissociates in water
The value of Ka for acetylsalicylic acid (aspirin) is 3.00×10−4. What is the value of Kb for its conjugate base, C9H7O4−? Dissociation Constant: The acid dissociation comes into the picture when a solution contains a weak acid, and it...
以NaOH滴定H3PO4 (Ka1=7.5×10-3,Ka2=6.2×10-8,Ka3=5.0×10-13)至生成Na2HPO4,溶液的pH值应当是( )。A.7.7B.8.7C.9.8D.10.7的答案是什么.用刷刷题APP,拍照搜索答疑.刷刷题(shuashuati.com)是专业的大学职业搜题找答案,刷题练习的工具.一键将文档转化为在线题库手机刷题,以
以NaOH 滴定H3PO4(Ka1=7.5×10-3,Ka2=6.2×10-8,Ka3=5.0×10-13)至生成Na2HPO4 时,A.的 pH 应当是( )。B.4.33C.12.3D.9.75E.7.21的答案是什么.用刷刷题APP,拍照搜索答疑.刷刷题(shuashuati.com)是专业的大学职业搜题找答案,刷题练习的工具.一键将文档转化为在线题库手机
Calculate the pH of 0.100 L of a buffer solution that is 0.21 M in HF (Ka = 3.5 x 10^-4) and 0.45 M in NaF. Calculate the change in pH when 2.0 x10-2mol of NaOH is added to 0.50 L of a buffer solution...
After adding 12.5g of NaOH to this solution, the pH was 6.03. What are the values of Ka1 and Ka2? Mr of Malonic Acid = 104g/mol19.5g of Malonic Acid = 19/104 = 0.1857molesmoles of NaOH added = 12.5/40 = 0.3175moles of acidic hydrogen available = 0.1857X2 = 0.3715moles...
Using the optimum conditions thus established (viz., 100g NaOH L-1, 30°C, and 30min) raised the high heating value (HHV) to 19.151MJkg-1 (i.e., by 4% relative to the starting material). Also, it allowed the content in elemental C to be preserved, that in H increased by 4.86...