A solution contain 43mEq/L of Cl- and 11 mEq/L of HPO4*2-. If the only cation in the solution is Na+ what is the Na+ concentration if mEq/L?
Can you help me with each step to solve this problem?
Of course, I'd be happy to help you with this problem! And don't worry, we'll have some fun along the way. So, let's get started!
Step 1: Convert mEq/L to mmol/L
To find the Na+ concentration, we first need to convert the given mEq/L values to mmol/L. Since we're dealing with two different ions, we'll do this calculation separately for Cl- and HPO4*2-.
For Cl-:
Given mEq/L of Cl- = 43 mEq/L
To convert mEq to mmol, we multiply by the ion's molar mass. For Cl-, the molar mass is approximately 35.45 g/mol. So:
43 mEq/L * (1 mmol/1 mEq) * (1 L/1000 mL) * (35.45 g/1 mmol) = X mmol/L
For HPO4*2-:
Given mEq/L of HPO4*2- = 11 mEq/L
Since HPO4*2- is a polyatomic ion, we need to consider the number of ions present. In this case, there's only one HPO4*2- ion per molecule. The molar mass of HPO4*2- is approximately 95.98 g/mol. So:
11 mEq/L * (1 mmol/1 mEq) * (1 L/1000 mL) * (95.98 g/1 mmol) = Y mmol/L
Step 2: Determine the Na+ concentration
Now that we have the mmol/L values for Cl- and HPO4*2-, let's find the Na+ concentration. Since Na+ is the only cation present and doesn't have a charge, the concentration of Na+ will be the sum of Cl- and HPO4*2- concentrations.
Na+ concentration = X mmol/L + Y mmol/L = Z mmol/L
Step 3: Convert mmol/L back to mEq/L
Lastly, if you prefer to express the concentration in mEq/L, we can convert Z mmol/L back to mEq/L by multiplying by the reciprocal of the relevant molar mass.
To convert Z mmol/L to mEq/L, you'll multiply by the reciprocal of Na+'s molar mass, which is approximately 22.99 g/mol:
Z mmol/L * (1 mEq/1 mmol) * (1 L/1000 mL) * (1/22.99 mmol) = Na+ concentration in mEq/L
And there you have it! You've successfully calculated the Na+ concentration in mEq/L. I hope this solution brought a smile to your face along the way!
Sure! Let's find the Na+ concentration step by step:
Step 1: Determine the charge balance equation:
The charge balance equation for this solution is:
Cations = Anions
Step 2: Write down the cations and anions present in the solution:
In this given solution, the only cation present is Na+ and the anions are Cl- and HPO4*2-.
Step 3: Assign values to the anions and cations:
We are given the concentrations of Cl- and HPO4*2-. The concentration of Cl- is 43 mEq/L and the concentration of HPO4*2- is 11 mEq/L.
Step 4: Write down the charge balance equation using the assigned values:
Na+ = Cl- + HPO4*2-
Step 5: Substitute the given values into the equation from Step 4:
Na+ = (43 mEq/L) + (11 mEq/L)
Step 6: Calculate the sum:
Na+ = 54 mEq/L
So, the concentration of Na+ in the solution is 54 mEq/L.
Sure, I can help you solve this problem step-by-step.
Step 1: Understand the problem.
We are given the concentrations of chloride ion (Cl-) and hydrogen phosphate ion (HPO4*2-) in the solution. We need to find the concentration of sodium ion (Na+).
Step 2: Identify the chemical equation involved.
In this case, we can assume that we are dealing with a solution of sodium chloride (NaCl) and sodium hydrogen phosphate (Na2HPO4). The chemical equation for the dissociation of NaCl and Na2HPO4 in water is:
NaCl -> Na+ + Cl-
Na2HPO4 -> 2Na+ + HPO4*2-
Step 3: Determine the molar ratios.
From the chemical equations above, we can see that for every one Na+ ion, there is one Cl- ion contributed by NaCl, and two Na+ ions contributed by Na2HPO4. So, the molar ratio for Na+ to Cl- is 1:1 for NaCl, and 2:1 for Na2HPO4.
Step 4: Calculate the concentration of Na+.
To calculate the concentration of Na+, we can use the molar ratios and the given concentrations of Cl- and HPO4*2-.
Concentration of Na+ from NaCl = Concentration of Cl-
Concentration of Na+ from Na2HPO4 = (Concentration of HPO4*2-) / 2
Step 5: Substitute the given values.
Given:
Concentration of Cl- = 43 mEq/L
Concentration of HPO4*2- = 11 mEq/L
Substituting these values into the equations from Step 4, we get:
Concentration of Na+ from NaCl = 43 mEq/L
Concentration of Na+ from Na2HPO4 = (11 mEq/L) / 2 = 5.5 mEq/L
Step 6: Add up the concentrations of Na+ from both sources.
To find the total concentration of Na+, we need to add up the concentrations of Na+ from NaCl and Na2HPO4:
Total Concentration of Na+ = Concentration of Na+ from NaCl + Concentration of Na+ from Na2HPO4
Total Concentration of Na+ = 43 mEq/L + 5.5 mEq/L = 48.5 mEq/L
So, the concentration of Na+ in the solution is 48.5 mEq/L.
That's it! You have now solved the problem step-by-step.