Can someone help me to solve this Chemistry problem and explain how?
Step 1. Look at the nature of the reactants. You have a reaction of a stong base and weak acid. Step 2. Look at the desired pH. You want acidic pH. That means that the acid will be in excess. Also since we said in step 1 that we have a weak acid and a strong base we will be forming a buffer system. In order to get the most accurate solution we would need to set-up an ICE table. However we can get an approximate solution which most of the times will be the same as the exact using the Henderson-Hasselbalch equations (and assumptions). pH=pKa+ log( [conj.base]/[acid] ) Step 3. Write the equation in order to determine the stoichiometry HCOOH + NaOH- -> HCOONa +H2O So the stoichiometry is 1 HCOOH : 1NaOH: 1 HCOONa Step 4. Find the number of moles of each compound If we used Vb L of NaOH then we used mole NaOH= MbVb from step 3, we know that we produced mole HCOONa=mole NaOH, thus mole HCOONa= MbVb we had mole HCOOH=MaVa so after the reaction remained mole HCOOH= MaVa-MbVb Step 5. Convert moles to concentrations. The volume of the solution becomes V=Va+Vb [conj.base]= [HCOO-] =[HCOONa] = MbVb/(Va+Vb) [acid]= [HCOOH]= (MaVa-MbVb)/(Va+Vb) Step 6 Divide the concentrations and then substitute in the equation. [conj.base]/[acid]= MbVb / (MaVa-MbVb) so pH= pKa+log (MbVb / (MaVa-MbVb)) => MbVb / (MaVa-MbVb)=10^(pH-pKa) => 0.05*Vb/ (0.1*0.25-0.05Vb)= 10^(4.5-(-log(1.7*10^-4)) )=5.38 => 0.05 Vb= 0.1345-0.269Vb => Vb=0.1345/0.319 =0.422 L =422 mL So the answer is B (There is a slight deviation probably because of the rounding of the 10^(pH-pKa))
Can someone help show me the steps to solving this chemistry question? Thanks?
hmmm. sounds like a Zumdahl problem... if we let x = pre-1982 pennies.. then (1254 - x) = post 1982 pennies then... 3.09x + 2.51(1254-x) = 3501.4 x = 610 1254 - 610 = 644 *********** do you understand? mass pre82 pennies x # pre82 pennies + mass post82 x #post82 pennies = total mass
I need help with one question in chemistry! Please put the steps and calculate it.?
I first tried finding pressure of benzene which you said was mol fraction x vapor pressure I found the mo fraction by first finding moles of benzene which was 52.6/ 78.1..I got .67mols then doing the same for toluene which came out to .8476 mols . the mol fraction for benzene was (.673mols/ 1.5211) then I multiplied that by vapor pressure (95.1 which i converted to atm which was .125atm) so .44 * .125atm and i ended up with .055 atm..I dont know if I am doing this right.
Can someone help me solve a Chemistry problem step by step?
A 50.0-mL sample of sodium carbonate solution was found to have a mass of 50.225g. When the solution was heated to dryness, the mass of the Na2CO3 Residue was 2.799g. Calculate the (a) mass/mass percentage concentration, and (b) molar concentration of the solution.
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how many grams of solute are needed to make the following concentrations of solutions: 750 mL of 0.35M solution of BaSO4 (not an exponent, but a small number at the bottom right)
Chemistry help please! Can you take me step by step on how to solve these equations?
1. What is the change in enthalpy when 6.0 g of NO are burned in the presence of excess oxygen according to 2NO(g) + O2(g) → 2NO2(g). ΔH for the reaction is -114.6 kJ/mol. 2. Raffinose is a sugar found in beans, cabbage, and brussels sprouts. If 6,600,000 J of heat were released by the combustion of raffinose according to C18H32O16 + 18O2 → 18CO2 + 16H2O, how many moles of CO2were generated? ΔH = -2025.5 kJ/mol for the combustion of raffinose.
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1000 mg = 1 g, so 4500 mg = 4.5 g you can set the problem up through ratios: (1000 mg)/(1g) = (4500 mg)/(x grams) 1*4500 = 1000*x x = 4.5 the rest are all done in similar fashion: 1 cm = 10 mm, so 25 cm = 250 mm 1 kg = 100 dag, so .005 kg = .5 dag 1 m = 100 cm, so .075 m = 7.5 cm 1 g = 1000 mg, so 15 g = 15000 mg