questions
1. write the balanced chemical equation for the reaction of khp with naoh.
2. suppose your laboratory instructor inadvertently gave you a sample of khp contaminated with nacl to use in standardizing your naoh. how would this affect the molarity you calculated for your naoh solu - tion? justify your answer.
3. how many grams of naoh are needed to prepare 500 ml of 0.125 m naoh?
4. a solution of malonic acid, h₂c₃h₂o₄, was standardized by titration with 0.1000 m naoh solution. if 20.76 ml of the naoh solution is required to neutralize completely 12.95 ml of the malonic acid solu - tion, what is the molarity of the malonic acid solution?
h₂c₃h₂o₄ + 2naoh → na₂c₃h₂o₄ + 2h₂o

Question

Accepted Answer

### Question 1
# Explanation:
## Step1: Identify reactants and products
Reactants: $KHC_8H_4O_4$ (KHP) and $NaOH$. Product: $KNaC_8H_4O_4$ and $H_2O$
## Step2: Balance the equation
$$KHC_8H_4O_4+NaOH = KNaC_8H_4O_4 + H_2O$$

# Answer:
$$KHC_8H_4O_4+NaOH = KNaC_8H_4O_4 + H_2O$$

### Question 2
# Brief Explanations:
- $NaCl$ does not react with $NaOH$.
- Molarity formula: $M=\frac{n}{V}$, where $n$ (moles of KHP) is calculated from mass ($n = \frac{m}{M}$, $M$ is molar mass of KHP).
- Contaminated KHP has less KHP by mass. So, calculated $n$ (of KHP) is lower.
- Using $M_{NaOH}=\frac{n_{KHP}}{V_{NaOH}}$, with lower $n_{KHP}$, calculated $M_{NaOH}$ is lower.

# Answer:
The calculated molarity of $NaOH$ solution would be lower. Because $NaCl$ (non - reacting) in KHP sample means less actual KHP (reactant with $NaOH$). Using $M=\frac{n}{V}$ ($n$ of KHP is lower as mass of pure KHP is less than assumed from contaminated sample mass), so $M_{NaOH}$ (calculated) is lower.

### Question 3
# Explanation:
## Step1: Calculate moles of $NaOH$
Molarity formula $M=\frac{n}{V}$, rearranged $n = M\times V$. $V = 500\space mL=0.5\space L$, $M = 0.125\space M$
$$n=0.125\space mol/L\times0.5\space L = 0.0625\space mol$$
## Step2: Calculate mass of $NaOH$
Molar mass of $NaOH$, $M_{NaOH}=40\space g/mol$. Using $m=n\times M$
$$m = 0.0625\space mol\times40\space g/mol=2.5\space g$$

# Answer:
$2.5\space g$

### Question 4
# Explanation:
## Step1: Use mole - ratio from balanced equation
From $H_2C_3H_2O_4+2NaOH = Na_2C_3H_2O_4 + 2H_2O$, mole - ratio $n_{H_2C_3H_2O_4}:n_{NaOH}=1:2$, so $n_{H_2C_3H_2O_4}=\frac{1}{2}n_{NaOH}$
## Step2: Calculate moles of $NaOH$
Using $n = M\times V$, $M_{NaOH}=0.1000\space M$, $V_{NaOH}=20.76\space mL = 0.02076\space L$
$$n_{NaOH}=0.1000\space mol/L\times0.02076\space L=0.002076\space mol$$
## Step3: Calculate moles of $H_2C_3H_2O_4$
$$n_{H_2C_3H_2O_4}=\frac{1}{2}\times0.002076\space mol = 0.001038\space mol$$
## Step4: Calculate molarity of $H_2C_3H_2O_4$
$V_{H_2C_3H_2O_4}=12.95\space mL=0.01295\space L$, using $M=\frac{n}{V}$
$$M_{H_2C_3H_2O_4}=\frac{0.001038\space mol}{0.01295\space L}=0.0802\space M$$

# Answer:
$0.0802\space M$

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QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Identify reactants and products

Step2: Balance the equation

Step1: Calculate moles of \(NaOH\)

Step2: Calculate mass of \(NaOH\)

Answer:

Question 2