Answer 1 :
NaOH + CH3COOH ----> CH3COONa + H2O
M1V1(NaOH) = M2V2(CH3COOH)
0.200M*94.7mL = M2*25mL
M2 = 0.200M *94.7mL / 25mL
M2 = 0.758M
Molarity of CH3COOH = 0.758M
por Prt 2 Short Answer Item 3 A 25.0 mL sample of vinegar (dilute acetic acid,...
Part A What is the molarity of the acetic acid solution? The reaction is NaOH(aq) + CH3CO2H(aq) +CH, CO2Na(aq) +H2O(1) IVO AEV O O ? M Submit Request Answer A 25.0 mL sample of vinegar (dilute acetic acid, CH3CO2H) is titrated and found to react with 94.7 mL of 0.200 MNaOH.
A 25.0 mL sample of 0.293 M acetic acid solution is titrated with 37.5 mL of 0.195 M NaOH solution. What is the pH of this solution if K = 1.8 x 10-5? Final Answer: _______
6. Vinegar is 5.00% (m/v) acetic acid (CH3COOH). How many milliliters of vinegar would you need to react with an excess of baking soda (NaHCO3) in order to generate 10.0 L of carbon dioxide gas at 1.05 atm and a temperature of 25.0°C? 7. Shown below is the aqueous reaction of KNOX, HCl, and SnCl2. 2 KNO2 (aq) + 6 HCl(aq) + 2 SnCl2 (aq) → 2 SnCla (aq) + N20 (g) + 3 H20 (1) + 2 KCl (aq)...
Part B A 15.0 mL sample of 40 % (m/v) acetic acid (CH3COOH) solution is added to water to give final volume of 24 mL. Express your answer using two significant figures. EVO AXC + 0 = ? Cfinal % (m/v) Submit Request Answer Provide Feedback Next >
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A 25.0 mL sample of an unknown acid in water is titrated with 0.1 MNaOH (aq) and resulted in the following data. Note: phenolphthalein indicator and a pH probe were used to monitor pH Volume (mL) 0.1 M NaoH obse pH Observed colour 3.15 3.56 2.5 3.96 4.43 7.5 8.16 10 12.5 11.32 12.09 15.0 17.5 12.37 20.0 12.61 22.5 12.81 25.0 12.86 12.89 27.5 12.91 30.0 Based on the data...