3. it takes 38 ml of 0.75 m naoh solution to completely neutralize 155 ml of a sulfuric acid solution…

Explanation:

Step1: Write the balanced chemical equation

$2NaOH + H_2SO_4=Na_2SO_4 + 2H_2O$ From the equation, the mole - ratio of $NaOH$ to $H_2SO_4$ is $n_{NaOH}:n_{H_2SO_4}=2:1$.

Step2: Calculate the moles of $NaOH$

Use the formula $n = M\times V$, where $M$ is the molarity and $V$ is the volume in liters. $V_{NaOH}=38\ mL = 0.038\ L$, $M_{NaOH}=0.75\ M$ $n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.75\ mol/L\times0.038\ L = 0.0285\ mol$

Step3: Calculate the moles of $H_2SO_4$

Since $n_{NaOH}:n_{H_2SO_4}=2:1$, then $n_{H_2SO_4}=\frac{n_{NaOH}}{2}$ $n_{H_2SO_4}=\frac{0.0285\ mol}{2}=0.01425\ mol$

Step4: Calculate the molarity of $H_2SO_4$

$V_{H_2SO_4}=155\ mL = 0.155\ L$ Use the formula $M=\frac{n}{V}$, so $M_{H_2SO_4}=\frac{n_{H_2SO_4}}{V_{H_2SO_4}}$ $M_{H_2SO_4}=\frac{0.01425\ mol}{0.155\ L}=0.092\ M$

Answer:

$0.092\ M$