During the titration the following reaction occurs:\[ \text{HA + NaOH} \rightarrow \text{NaA + H}_2\text{O} \]- **What is the concentration of A⁻ (the conjugate base of the weak acid) at the equivalence point?** \[ \_\_\_\_\_ \text{ M} \]- **Use [A⁻] and the pH at the equivalence point to estimate Kₐ of the weak acid (this is the hard way of estimating Kₐ).** \[ \_\_\_\_\_ \]- **Use the pH at the half-way point to estimate Kₐ of the weak acid (this is the easier way).** \[ \_\_\_\_\_ \] A student performs a titration, titrating 25.00 mL of a weak monoprotic acid, HA, with a 1.18 M solution of NaOH. They collect data, plot a titration curve, and determine the values given in the table below.| | mL NaOH added | pH ||---------------|---------------|-----|| Half-way Point | 17.66 | 3.46|| Equivalence point | 35.31 | 8.49|**Questions:**1. How many moles of NaOH have been added at the equivalence point? ____ mol2. What is the total volume of the solution at the equivalence point? ____ mL

Hii there, As there are multiple sub parts posted. we are answering first-three sub parts . If you need further question to be answered. please repost the question with specific question number. Given: Volume of HA = 25.00 mL Concentration of NaOH = 1.18 M mL NaOH added pH Half-way Point 17.66 3.46 Equivalence Point 35.31 8.49 Calculation for moles of NaOH at Equivalence Point: moles=molarity×volumemoles of NaOH=1.18 mol/L×35.31×10-3 Lmoles of NaOH=0.0417 mol Calculation for total volume of solution at Equivalence Point: Total volume of solution=volume of HA+volume of NaOH at equivalence pointTotal volume of solution=25.0 mL+35.31 mLTotal volume of solution=60.31 mL Chemical reaction is given by, HA+NaOH⇌NaA+H2OHA+OH-⇌A-+H2O At equivalence point no. of moles of HA will be equal to that of NaOH. Thus the concentration of A-(conjugate base of HA) can be calculated as follows: moles of A-=moles of NaOHmoles of A-=0.0417 molConcentration of A-=moles of A-Volume of olutionConcentration of A-=0.0417 mol60.31×10-3 LConcentration of A-=0.691 M

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Chemistry

A student peforms a titration, titrating 25.00 mL of a weak monoprotic acid, HA, with a 1.18 M solution of NaOH. They collect data, plot a titration curve and determine the values given in the below table. ml NaOH added pH Half-way Point 17,66 3.46 Equivalence point 35.31 8.49 How many moles of NaOH have been added at the equivalence point? mol What is the total volume of the solution at the equivalence point? mL

A student peforms a titration, titrating 25.00 mL of a weak monoprotic acid, HA, with a 1.18 M solution of NaOH. They collect data, plot a titration curve and determine the values given in the below table. ml NaOH added pH Half-way Point 17,66 3.46 Equivalence point 35.31 8.49 How many moles of NaOH have been added at the equivalence point? mol What is the total volume of the solution at the equivalence point? mL

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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