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How Many milliliters of 0.258 M NaOH are required to complétely neutrilize 2.00 g of aceitic acid HC2H3O2?
To find the volume of 0.258 M NaOH needed to neutralize 2.00 g of acetic acid (HC2H3O2), you can use the molar ratio between NaOH and acetic acid. First, determine the moles of acetic acid using its molar mass. Then, use the mole ratio from the balanced chemical equation to find moles of NaOH required. Finally, use the concentration of NaOH to find the volume needed.
Balanced equation of HCH3COO NaOH?
The balanced equation for the reaction between acetic acid (HC2H3O2) and sodium hydroxide (NaOH) is: HC2H3O2 + NaOH → NaC2H3O2 + H2O
40.00 mL of 0.10 M HC2H3O2 is titrated with 0.15 M of NaOH. Ka of HC2H3O2 is 1.8x10-5 What volume of NaOH is used to reach equivalence pt and what molar concentration of C2H3O2- equivalence pt?
40.00(mL)*0.10(M) = VNaOH(mL)*0.15(M) , soVNaOH(mL) = 40.00*0.10 / 0.15 = 26.67 = 27 mLCacetate = 0.10 * [ 40.00 / (40 mL + 26.7 mL) ] = 0.060 MMark: nothing to do with Kacid !
What volume of NaOH was used in the titration?
To determine the volume of NaOH used in the titration, you need to know the concentration of the NaOH solution and the volume required to reach the endpoint. Use the formula: volume NaOH (L) = volume HCl (L) * concentration HCl / concentration NaOH.
What is the maximum volume of 0.10 m naoh that can be completely neutralized by 0.20 ml of hcl?
To determine the maximum volume of 0.10 M NaOH that can be neutralized by 0.20 Ml of HCl, you need to use the equation: moles = Molarity × Volume. First, calculate the moles of HCl used (0.20 ml * 0.20 M) and then use the mole ratio from the balanced chemical equation to determine the moles of NaOH needed. Finally, divide the moles of NaOH by the concentration of NaOH to find the volume that can be neutralized.
What is the balanced equation for the reaction of acetic acid and NaOH?
The balanced chemical equation for the reaction between acetic acid (HC2H3O2) and sodium hydroxide (NaOH) is: HC2H3O2 + NaOH → NaC2H3O2 + H2O This reaction forms sodium acetate (NaC2H3O2) and water (H2O).
A student added 100 ml of 0.10 m naoh to 250 ml of a buffer that contained 0.15 m hc2h3o2 and 0.25 m c2h3o2- what is the concentration of hc2h3o2 after the addition of the strong base?
CHEM 123 Solution to Ques 16.65 from Wiley #12 16.65 A student added 100 mL of 0.10 M NaOH to 250 mL of a buffer that contained 0.15 M HC2H3O2 and 0.25 M C2H3O2−. What is the concentration of HC2H3O2 after the addition of the strong base. Begin by writing the equation for the buffer (acid dissociation of the weak acid) HC2H3O2 + H2O H3O + + C2H3O2- Ka = 1.8 × 10-5 Initial amounts in the solution are: # mol C2H3O2- = (0.25 mol/L)(0.25 L) = 0.063 mol C2H3O2- # mol HC2H3O2 = (0.15 mol/L)(0.25 L) = 0.038 mol HC2H3O2 The added base (0.100 L)(0.10 mol/L) = 0.010 mol NaOH) will react with the acetic acid present in the buffer solution: NaOH + HC2H3O2 ⎯→ H2O + NaC2H3O2 Assume the added base reacts completely (since it's a strong base with a weak acid). For each mole of base added, one mole of HC2H3O2 is converted to C2H3O2-. Since 0.010 mol of base is added, 0.010 mol of HC2H3O2 will be depleted from the initial amount (0.038 mol) and 0.010 mol C2H3O2− ADDITIONAL produced in this reaction (added to the initial amount of 0.063 mol C2H3O2−): # mol HC2H3O2final = (0.038 - 0.010) mol = 0.028 mol # mol C2H3O2-final = (0.063 + 0.010) mol = 0.073 mol The final volume of solution is 250 mL + 100 mL = 350 mL. AFTER the reaction with NaOH, the concentrations become… [HC2H3O2] = 0.028 mol/0.350L = 0.080 M HC2H3O2 [C2H3O2−] = 0.073 mol/0.350L = 0.21 M C2H3O2− In Wiley #12, the question asked only what the concentration of acetic acid becomes, so this would be the answer: 0.080 M. (The optional given you in Wiley was 0.079 M.) In the actual question in the book, it went further and asked you what is the CHANGE in concentration of HC2H3O2 and C2H3O2−, so you would simply go one further step: Initial conc of HC2H3O2 was 0.15M Change in conc = 0.15−0.080 = 0.0.07 M decrease in [HC2H3O2] Initial conc of C2H3O2− was 0.25M Change in conc = 0.25M − 0.21 Μ = 0.04 Μ decrease in [C2H3O2−]
A 25.00ml sample of h2so4 requires 22.65 ml of the 0.550m naoh for its titration what was the concentration of sulfuric acid?
The balanced chemical equation for the reaction is H2SO4 + 2NaOH -> Na2SO4 + 2H2O. From the mole ratio, 1 mole of H2SO4 reacts with 2 moles of NaOH. Using the volume and concentration of NaOH, we can calculate the moles of NaOH used. Then, knowing the moles of NaOH used and the volume of H2SO4, we can find the concentration of sulfuric acid.
How many liters of naoh are in 1.3 moles of 3.42 m of naoh?
Molarity = moles of solute/Liters of solution 3.42 M NaOH = 1.3 moles NaOH/Liters NaOH Liters NaOH = 1.3 moles NaOH/3.42 M NaOH = 0.38 Liters
How many ml of 250 M NaOH are required to neutralize 30.4 ml of 152 M HCl?
To neutralize the acid, we need to use the same number of moles of base. First, calculate the number of moles of HCl using its concentration and volume. Then, use the mole ratio from the balanced equation to find the required volume of NaOH. Convert the volume to mL.
What is the ratio of moles of CuSO4 to moles of NaOH?
This ratio is 1:2.
How many grams of solid NaOH are needed to prepare 500mL NaOH solution with a concentration of 0.4M?
molarity equals moles of solute /volume of solution in litres . moles of NaOH equals 5g/40g = 0.125 and volume of solution will be volume of water + volume of NaOH = 0.5 litre+0.002 l which is nearly 0.5 litre . (volume of NaOH is calculated by its density) so molarity = 0.125mol/0.5litre = 0.25 M
