Calculate the number of milliliters of 0.754 M KOH required to precipitate all of the Zn2+ ions in 162 mL of 0.731 M Zn(NO3)2 solution as Zn(OH)2. The equation for the reaction is: Zn(NO3)2(aq) + 2KOH(aq) Zn(OH)2(s) + 2KNO3(aq)

A 0.5504-g sample of KHP (potassium hydrogen phthalate, KHC8H4O4; molar mass = 204.22 g/mol) is completely dissolved in enough deionized water to make 50.00 mL of solution. The titration of this solution with a KOH solution requires 20.50 mL of the base (KOH solution) to reach the “end-point”. What is the molar concentration of KOH solution? The reaction occurs as follows:                       KOH(aq)  +  KHC8H4O4(aq)  -->  K2C8H4O4(aq)  +  H2O(l)           (A) 0.2457 M           (B) 0.1315 M           (C) 0.05390 M           (D) 0.03823 M

A 0.450 g sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq). The equation for the reaction is CaCO3(s) + 2HCl(aq) - CaCl2(aq) + H2O(l) + CO2(g) The excess HCl(aq) is titrated by 5.60 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO3(s) in the sample. mass percentage:

If 785 mL of 0.350 M AgNO₃ is added to 475 mL of 0.250 M CaBr₂, how many moles of AgBr precipitate will be formed? The balanced equation is 2 AgNO₃(aq) + CaBr₂(aq) → 2 AgBr(s) + Ca(NO₃)₂(aq)

Calculate the number of milliliters of 0.580 M Ba(OH)2 required to precipitate all of the Cu2+ ions in 120 mL of 0.745 M CuCl2 solution as Cu(OH)2. The equation for the reaction is:CuCl2(aq) + Ba(OH)2(aq) Cu(OH)2(s) + BaCl2(aq) mL Ba(OH)2

A solution of Ca(OH)2 (aq) of unknown concentration is titrated to the equivalence point with 0.250 M HCl(aq). The reading on the acid buret is 11.25 mL at the first equivalence point.  The reading on the Base buret at the equivalence point is 10.10 mL. What is the concentration (molarity) of the Ca(OH)2 solution?

A student used a pH meter to collect data for the titration of an unknown concentration of propanoic acid with a 0.150 mol/L solution of sodium hydroxide. The data table from the investigation is shown in the table below. Data for the Titration of Propanoic Acid with Sodium Hydroxide Volume of propanoic acid: 25.00 mL [NaOH] = 0.150 mol/L Volume of NaOH (aq) added (mL) 0.00 2.00 4.11 7.98 11.95 14.08 16.05 17.00 17.21 17.39 17.62 17.99 18.18 18.39 18.80 20.00 22.03 PH 2.83 3.84 4.20 4.64 5.03 5.27 5.61 5.90 6.00 6.09 6.23 6.74 8.80 10.92 12.24 12.56 12.69 Adobe 4343625416 00..

By titration, it is found that 62.3 mL of 0.194 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentration of the HCl solution.[HCl]= M