Question

1. it is required to consume 25~30 ml of a 0.1 mol·l⁻¹ naoh solution during titration. how many grams of the primary standard potassium hydrogen phthalate (khc₈h₄o₄) should be weighed? if oxalic acid dihydrate (h₂c₂o₄·2h₂o) is used as the primary standard instead, how many grams should be weighed? if the weighing error of the balance is ±0.2 mg, what are the relative errors of the weighing of the above two reagents? what does the calculation result indicate? m(khc₈h₄o₄) = 204.22 g·mol⁻¹; m(h₂c₂o₄·2h₂o) = 126.07 g·mol⁻¹ for khc₈h₄o₄:the mole ratio of khc₈h₄o₄ to naoh is 1:1. for h₂c₂o₄·2h₂o (regarded as a diprotic acid, the mole ratio to naoh is 1:2

Explanation:

Step1: Calculate the amount of NaOH

For the volume range of NaOH solution $V_{NaOH}=25 - 30\ mL=0.025 - 0.030\ L$ and concentration $c_{NaOH}=0.1\ mol/L$. The amount of substance of NaOH is $n_{NaOH}=c_{NaOH}\times V_{NaOH}$.
When $V_{NaOH} = 0.025\ L$, $n_{NaOH1}=0.1\ mol/L\times0.025\ L = 0.0025\ mol$; when $V_{NaOH}=0.030\ L$, $n_{NaOH2}=0.1\ mol/L\times0.030\ L=0.003\ mol$.

Step2: Calculate the mass of $KHC_{8}H_{4}O_{4}$

Since the mole - ratio of $KHC_{8}H_{4}O_{4}$ to $NaOH$ is $1:1$, $n_{KHC_{8}H_{4}O_{4}}=n_{NaOH}$.
The molar mass of $KHC_{8}H_{4}O_{4}$ is $M = 204.22\ g/mol$.
When $n_{KHC_{8}H_{4}O_{4}}=n_{NaOH1}=0.0025\ mol$, $m_{1}=n_{KHC_{8}H_{4}O_{4}}\times M=0.0025\ mol\times204.22\ g/mol = 0.51055\ g$; when $n_{KHC_{8}H_{4}O_{4}}=n_{NaOH2}=0.003\ mol$, $m_{2}=0.003\ mol\times204.22\ g/mol = 0.61266\ g$.

Step3: Calculate the relative error of weighing $KHC_{8}H_{4}O_{4}$

The weighing error $\Delta m=\pm0.2\ mg = \pm0.0002\ g$.
When $m = 0.51055\ g$, the relative error $RE_{1}=\frac{\Delta m}{m}=\frac{0.0002\ g}{0.51055\ g}\approx0.04\%$; when $m = 0.61266\ g$, the relative error $RE_{2}=\frac{0.0002\ g}{0.61266\ g}\approx0.03\%$.

Step4: Calculate the mass of $H_{2}C_{2}O_{4}\cdot2H_{2}O$

Since the mole - ratio of $H_{2}C_{2}O_{4}\cdot2H_{2}O$ to $NaOH$ is $1:2$, $n_{H_{2}C_{2}O_{4}\cdot2H_{2}O}=\frac{1}{2}n_{NaOH}$.
The molar mass of $H_{2}C_{2}O_{4}\cdot2H_{2}O$ is $M = 126.07\ g/mol$.
When $n_{NaOH}=n_{NaOH1}=0.0025\ mol$, $n_{H_{2}C_{2}O_{4}\cdot2H_{2}O1}=\frac{1}{2}\times0.0025\ mol = 0.00125\ mol$, $m_{3}=n_{H_{2}C_{2}O_{4}\cdot2H_{2}O1}\times M=0.00125\ mol\times126.07\ g/mol=0.1575875\ g$; when $n_{NaOH}=n_{NaOH2}=0.003\ mol$, $n_{H_{2}C_{2}O_{4}\cdot2H_{2}O2}=\frac{1}{2}\times0.003\ mol = 0.0015\ mol$, $m_{4}=0.0015\ mol\times126.07\ g/mol = 0.189105\ g$.

Step5: Calculate the relative error of weighing $H_{2}C_{2}O_{4}\cdot2H_{2}O$

When $m = 0.1575875\ g$, the relative error $RE_{3}=\frac{\Delta m}{m}=\frac{0.0002\ g}{0.1575875\ g}\approx0.13\%$; when $m = 0.189105\ g$, the relative error $RE_{4}=\frac{0.0002\ g}{0.189105\ g}\approx0.11\%$.

The calculation result indicates that for the same weighing error, the relative error of weighing the reagent with a larger molar - mass and a larger amount of substance (under the same reaction condition) is smaller.

Answer:

The mass of $KHC_{8}H_{4}O_{4}$ should be $0.51 - 0.61\ g$, and the relative errors of weighing are about $0.03 - 0.04\%$. The mass of $H_{2}C_{2}O_{4}\cdot2H_{2}O$ should be $0.16 - 0.19\ g$, and the relative errors of weighing are about $0.11 - 0.13\%$.