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If 0.007 g of KHP (molar mass = 204.23) requires 30.0 mL of NaOH upon titration, what is the normality of NaOH?
Normality (N) = m /V × 1/eq= 0.007/0.03 × 1/Eq= 39.997 × 0.2333= 9.333
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How are you going to prepare 1.5 L of 0.1N KOH solution? Molar mass of KOH = 56 g/mol.
1.5 L of 0.1N KOH solution Have moles of KOH = 1.5 * 0.1 = 0.15 mol . Mass of KOH = 0.15 * 56 = 8.4 g . Answer...
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23.91 mL of a standard solution of KHP of molarity 0.1454M used to standardize a solution of NaOH giving final volume of 26.31 mL of NaOH at endpoint. Calculate concentration of NaOH
0.1321M
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Calculate the mass of potassium hydrogen phthalate (KHP) needed to prepare a 0.1 M solution in a 250.0 ml volumetric flask given that the molar mass of potassium hydrogen phthalate is 204.23 g/mol
m=204.23×250/1000×0.1=5.10575 (g)Mass of potassium hydrogen phthalate (KHP) needed is 5.1058 g.
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1 g of NaOH was added to 30 cm3 of 1 mol/dm3 HCl. How many cubic meters 0.1 mol/dm3 KOH solution will neutralize excess acid?
The chemical equation is HCl + NaOH = NaCl + H2O n(NaOH) = m(NaOH)/MW(NaOH) = 1/(23 +16 +1) = 0.025 mol 30 cm3 = 30 mL = 0.030 L 1mol/dm3...
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A sample weighing 4.52g containing KHP (MM=204.2 g/mol) required 28 ml of 0.15M NaOH solution for neutralization, calculate the mass percent of KHP in this sample
Solution:KHP = potassium hydrogen phthalate = KHC8H4O4The balanced chemical equation:NaOH(aq) + KHP(aq) = NaKP(aq) + H2O(l)According to the chemical equation: n(NaOH) = n(KHP)Moles of NaOH = n(NaOH) = Molarity of...
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Calculate the concentration of OH- ions when reacting with 100 cm3 of KOH solution, concentration 0.09 mol / dm3 with 120 cm3 of H2SO4 solution with concentration 0.1 mol / dm3!
The reaction given:2KOH + H2SO4 = K2SO4 + 2H2OAs 1 L = 1dm3 and 1 ml = 1 cm3, the number of moles of the reactants equals:n(KOH) = M × V = 0.09...
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From an initial mix of: 2 mol Cl2, 1 mol Br2, 0.5 mol I2, 0.5 mol Cl−, 1 mol Br−, 2 mol I−; the total quantity of I2 present after all possible reactions occur is
0.5 mol I2 and 0.5 mol Cl- will not react with anything; 2 mol I- will react with one mol of Cl2 producing one mol of I2: 2I- + Cl2...
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What volume of a 0.537 M barium hydroxide solution is needed to exactly neutralize 7.57 grams of KHP ?
CM (Ba(OH)2)=0.537 M m(KHP)=7.57 g V(Ba(OH)2 )- ? Ba(OH)2 + 2KHP = Ba(KP)2 +2H2O n(KHP)= m(KHP)/M(KHP)=7.57g / 71g/mol= 0.106mol. n(Ba(OH)2)=n(KHP)/2=0.106mol/2= 0.053mol. CM (Ba(OH)2)= n(Ba(OH)2)/ V(Ba(OH)2 ). V(Ba(OH)2 )= n(Ba(OH)2)/ CM...
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34.62 mL of a 0.3 M Mg(OH)2 solution are needed to neutralize a 15-mL solution of HCl on unknown concentration. Calculate the concentration of HCl solution.
Mg(OH)2+2HCl=MgCl2+2H2O1.Trace by finding the amount of Mg(OH)2. To do this, we multiply the molarity by the volume (liters): 0.03462litres *0.3 mol/litres= 0.010386 mol Mg(OH)22.based on the reaction we find the amount...
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