An analytical chemist is titrating 179.6 mL of a 0.8400M solution of cyanic acid (HCNO) with a 1.200M solution of KOH. The p K, of cyanic acid is 3.46. Calculate the pH of the acid solution after the chemist has added 144.3 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places. 0 pH =

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A: Given: Volume of HCNO = 216.4 mL Concentration of HCNO = 0.480 M Volume of NaOH = 363.3 mL…

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A: Given -> Volume of NH3 = 178.6 ml Molarity of NH3 = 0.7500 M Volume of HNO3 = 142.1 ml Molarity…

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A: By the reaction of Propanoic acid and KOH potassium propanoate and salt.sodium propanoate and…

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A: Given that, Volume of 0.6200 M HN3 acid = 163.7 mL  = 0.1637 L Moles of HN3 acid = 0.62 M×0.1637 L =…

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A: Henderson–Hasselbalch equation, pH = pKa + log [salt]/[acid]

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A: no of moles of C3H7NH2  = molarity * volume in L                                      =…

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A: To Calculate the pH, we would first Calculate moles of both compounds and draw an ICE table.  Based…

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A: The volume of KOH solution required to neutralize the given volume of formic acid is calculated…

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A: The reaction is as follows:

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A:

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Q: A solution of the weak acid HA is prepared by dissolving 2.20 g of HA in 100.0 mL water. The…

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A: From given  Initially we are calculating moles of nitric acid and propylamine individually and then…

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A: The given data is as follows: The concentration of propionic acid = 0.6500 M The volume of propionic…

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Q: An analytical chemist is titrating 216.5 mL of a 0.8400 M solution of dimethylamine ((CH,) NH) with…

A: We have,    Thus,

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A:

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A: 239.5 mL of a 1.000M solution of propionic acid (HC2H5CO2) is titrated with a 0.2500M solution of…

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A: Moles of HN3 = Molarity × Volume                         = 0.5200 M × 240.8 ml = 125.216 mmol Moles…