an naoh solution of unknown concentration is titrated with hcl. 33.0 ml of 0.250 mol/l hcl will neutralize…

Explanation:

Step1: Write the balanced chemical equation

$$\text{NaOH}+\text{HCl}=\text{NaCl}+\text{H}_2\text{O}$$ From the equation, the mole ratio of $\text{NaOH}$ to $\text{HCl}$ is $1:1$.

Step2: Calculate the moles of $\text{HCl}$

Use the formula $n = c\times V$ (where $n$ is moles, $c$ is concentration, and $V$ is volume in liters). $V_{\text{HCl}}=33.0\space mL = 0.033\space L$, $c_{\text{HCl}} = 0.250\space mol/L$ $n_{\text{HCl}}=c_{\text{HCl}}\times V_{\text{HCl}}=0.250\space mol/L\times0.033\space L = 0.00825\space mol$

Step3: Determine the moles of $\text{NaOH}$

Since the mole ratio of $\text{NaOH}$ to $\text{HCl}$ is $1:1$, $n_{\text{NaOH}}=n_{\text{HCl}} = 0.00825\space mol$

Step4: Calculate the concentration of $\text{NaOH}$

$V_{\text{NaOH}}=50.0\space mL=0.050\space L$ Use the formula $c=\frac{n}{V}$ $c_{\text{NaOH}}=\frac{n_{\text{NaOH}}}{V_{\text{NaOH}}}=\frac{0.00825\space mol}{0.050\space L}=0.165\space mol/L$

Answer:

$0.165\space mol/L$